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ECLECTIC    EDUCATIONAL    SERIES. 


RAY'S  ARITHMETIC,  THIRD  BOOK 


PRACTICAL 


ARITHMETIC, 


BY 


INDUCTION  AND   ANALYSIS. 


By    JOSEPH  [RAY,    M.D. 

LATE   PEOFESSOE    OF    MATHEMATICS    IN    WOODWAED    COLLEGE. 


ONE  THOUSANDTH  EDITION— IMPKOVED. 


VAN  ANTWERP,  BRAGG  &  CO., 

CINCINNATI:  NEW   YORK: 

137  WALNUT  ST.  28  BONO  ST. 


roTun 

SPECIAL    NOTICE. 

Ray's  Arithmetics  have  recently  been  thoroughly  revised, 
and  issued  as — 

RAY'S   NEW  ARITHMETICS. 

Ba^»  New  Primai-y  Arithmetic. 
Bay's  New  InteUectucU  Arithmetic 
Hay's  New  Practical  Aj'ithmetic. 


6b4  1 


Ray's  Two-Book  Series. 

Hay's  New  Elementai-y  Arithmetic. 
Bay's  New  Practical  Arithmetic. 

For  High  Schools  and  Colleges. 
Bay's  New  Higher  Arithmetic 

The  many  changes  in  business  transactions,  as  well  aa 
the  advance  in  methods  of  instruction,  have  made  such  re- 
vision necessary.  The  New  Arithmetics  are  sold  for  the  same 
low  prices  as  the  old  editions,  notwithstanding  the  paper,  print- 
ing, binding,  and  general  appearance  are  far  superior.  Special 
terms  for  exchange  of  the  new  series  for  the  old,  can  be  had 
by  application  to  the  publishers. 

VAN  ANTWERP,  BRAGG  &  CO. 

/OZ 

^^-.^ 

Ki)t<!ved  iicconlin;^  to  Act  of  Congress,  in  the  vear  KiBhteen  Hundred 
and  Fifty-Seven,  by  WiNXHrtoi'  B.  Smiiit,  in  the  Clerk's  OfHce  of  the 
District  Court  of  tlie  United  States,  for  the  Southern  District  of  Ohio. 

Copyright, 

1885, 

BY  Van  Antwerp,  Braqg  &  Co. 


PREFACE. 

Few,  if  any,  works  on  Arithmetic,  have  received  the  approbation 
which  has  been  bestowed  upon  "Rays  Arithmetic,  Part  Third;" 
and  the  constantly  increasing  demand  for  it  having  rendered 
another  renewal  of  the  stereotype  plates  necessary,  it  has  been 
made  the  occasion  for  remodeling  and  greatly  improving  the 
volume  in  all  its  parts. 

The  Inductive  and  Analytic  methods  here  adopted,  lead  the 
learner  to  an  understanding  of  the  principles  from  which  the 
rules  are  derived,  and  teach  him  to  regard  rules  as  results  rather 
than  reasons :  he  will  understand  the  *'  why  and  wherefore "  of 
every  operation  performed,  and  gain  a  thorough  knowledge  of 
the  principles  of  Arithmetic  with  its  applications,  while  the 
faculties  of  the  mind  are  strengthened  and  disciplined. 

The  leading  features  of  the  work  are : 

1st.  Every  principle  is  clearly  explained  by  an  analysis  or 
solution  of  simple  examples,  from  which  a  Rule  is  derived.  This 
is  followed  by  graduated  exercises  designed  to  render  the  pupil 
familiar  with  its  application. 

2d.  The  arrangement  is  strictly  philosophical;  no  principle  is 
anticipated:  the  pupil  is  never  required  to  perform  any  operation 
until  the  principle  on  which  it  is  founded  has  first  been  explained. 
For  this  reason,  those  processes  of  reduction  that  require  the  use 
of  fractions,  are  introduced  after  fractions. 

3d.  The  subject  of  Fractions,  a  thorough  understanding  of 
which  is  almost  a  knowledge  of  Arithmetic,  has  received  that 
attention  which  its  use  and  importance  demand. 

4th.  The  subject  of  Proportion  is  introduced  immediately  after 
Decimals;  this  enables  the  instructor  to  treat  Percentage  and  its 
various  applications,  either  by  proportion,  or  by  analysis,  as  he 
may  prefer. 


4  PREFACE. 

5th.  Particular  attention  has  been  given  to  render  the  work 
practical;  the  weiglits  and  measures  are  referred  to,  and  con- 
form to  the  legal  standards;  while  pounds,  shillings,  and  pence, 
being  no  longer  used  in  actual  business,  are  only  introduced 
under  Exchange. 

While  Federal  Money  may  be  considered  in  connection  with 
decimals,  yet  it  is  truly  a  species  of  compound  numbeis,  and  w 
so  regarded  in  all  the  ordinary  computations  of  business.  Hence, 
the  propriety  of  assigning  it  the  place  which  it  occupies  in  this 
work. 

While  cancellation  is  introduced,  it  is  not  made  a  hobby,  or  an 
arithmetical  machine,  by  which  results  can  be  obtained  merely 
in  a  mechanical  manner. 

The  object  throughout  has  been  to  combine  practical  utility  with 
scientific  accuracy;  to  present  a  work  embracing  the  best  methods, 
with  all  real  improvements.  How  far  this  object  has  been 
secured,  is  submitted  to  those  engaged  in  the  laborious  and 
responsible  work  of  education. 

UiG^  An  Appendix  has  been  added  to  this  edition,  presenting 
the  Metrical  System  of  Weights  and  Measures.  This  has  been  com- 
piled chiefly  from  modern  French  works  on  this  subject;  and  was 
submitted,  in  manuscript,  to  the  critical  examination  of  Pbof. 
H.  A.  Newton,  of  Yale  College,  to  whom  we  are  under  great  obli- 
gations for  such  corrections,  additions,  and  emendations  as  his 
intimate  knowledge  of  the  subject  suggested. 


A  New  Book.— Ray's  Test  Examples. 

Three  Tliousand  Test  Examples  in  Arithmetic;  Prac- 
tical Problems  for  the  Slate  or  Blackboard,  for  Drill 
exercises  and  Review. 

Designed  to  save  the  teacher  much  time  and  labor  by  furnishing, 
ready  to  his  hand,  a  largo  number  and  variety  of  Drill  Exer- 
cises for  frequent  and  thorough  review. 

Two  Editions  :  With  Answers, — Without  Answers. 


CONTENTS. 


Simple  Numbers.  pace. 

Notation  and  Numeration, 9 

Addition, 19 

Subtraction, 26 

Multiplication, 33 

Division, 43 

Eeview  of  Principles, 61 

Promiscuous  Examples, 63 

General  Principles  of  Division, 65 

Cancellation, 67 

Compound  Numbers. 

Definitions, 70 

United  States  or  Federal  Money, 71 

Notation  and  Numeration  of  U.  S.  Money, 72 

Eeduction  of  U.  S.  Money, 73 

Addition  of  U.  S.  Money, 75 

Subtraction  of  U.  S.  Money, 76 

Multiplication  of  U.  S,  Money, 77 

Division  of  U.  S.  Money, 78 

Promiscuous  Examples  in  U.  S.  Money, 81 

Eeduction  of  Compound  Numbers, 83 

Dry  Measure, 83 

Troy  or  Mint  Weight, 87 

Apothecaries  Weight, 88 

Avoirdupois  Weight, 89 

Long  or  Linear  Measure, 90 

Land  or  Square  Measure, 90 

To  find  the  Area  of  a  Eectangle, 92 

Solid  or  Cubic  Measure, 94 

Cloth  Measure, 95 

Wine  or  Liquid  Measure, 96 

Ale  or  Beer  Measure, 97 

Time  Measure, 97 

Circular  Measure, 99 

Miscellaneous  Table,.. 99 

Promiscuous  Examples  in  Eeduction, 100 

Addition  of  Compound  Numbers, 10.3 

Subtraction  of  Compound  Numbers, 107 

To  find  the  period  of  Time  between  any  two  Dates, 110 

Multiplication  of  Compound  Numbers, Ill 

Division  of  Compound  Numbers, 114 

Longitude  and  Time, 117 


(5  CONTENTS. 

Factoring.  page. 

Definitions, 120 

Six  Propositions, 122 

E.xamples  for  Practice, 125 

GuEATEST  Common  Divisor. 

First  Method,  by  Factoring, 127 

Second  Method,  by  Division, 128 

Least  Common  Multiple. 

First  Method,  by  Factoring, 131 

Second  Method,  by  Division, 182 

Common  Fractions. 

Origin  and  Nature  of  Fractions, 134 

Definitions, 138 

General  Principles, 139 

Reduction  of  Fractions, 142 

To  reduce  a  fraction  to  its  lowest  terms, 142 

To  reduce  an  improper  fraction  to  a  whole  or  mixed  number, 143 

To  reduce  a  whole  or  mixed  number  to  an  improper  fraction, 144 

To  reduce  compound  to  simple  fractions, 146 

To  reduce  compound  to  simple  fractions  by  Cancellation, 147 

To  reduce  fractions  to  a  common  denominator, 148 

To  reduce  fractions  to  the  least  common  denominator, 150 

Addition  of  Fractions, 152 

Subtraction  of  Fractions, 154 

Multiplication  of  Fractions, 156 

Division  of  Fractions, 162 

Fractional  Exercises  in  Compound  Numbers, 169 

Reduction  of  Fractional  Compound  Numbers, 171 

Addition  and  Subtraction  of  Fractional  Compound  Numbers, 174 

Promiscuous  Examples, 175 

Decimal  Fractions. 

Origin  and  nature  of  Decimals, 177 

Decimal  Numeration  and  Notation, 180 

Addition  of  Decimals, 183 

Subtraction  of  Decimals, 184 

Multiplication  of  Decimals, 185 

Division  of  Decimals, 187 

Reduction  of  Decimals, 189 

Promiscuous  Examples, 198 

Ratio. 

The  nature  of  Ratio, 195 

Mcthotl  of  expressing  Ratio, 196 

Propoution. 

The  naturo  of  Proportion, 197 

Simple  Proportion, 199 


CONTENTS.  7 

Proportion  Continued.  page. 

Kule  of  Cause  and  Eflfcct, 205 

Compound  Proportion, 206 

Aliquots,  OR  Practice 209 

Percentage  and  its  applications. 

To  find  any  per  cent,  of  a  Number, 212 

To  find  what  per  cent,  one  Number  is  of  another, 214 

Commission, 215 

Insurance, 217 

Stocks, 218 

Brokerage, 219 

Interest. 

Simple  Interest, 220 

General  Rule  for  Interest, 225 

Another  method  for  Interest, 227 

Partial  Payments, 229 

Problems  in  Interest, 233 

Compound  Interest,.., 236 

Discount, 239 

Profit  and  Loss, 245 

Assessment  of  Taxes, 250 

American  Duties, , 252 

Partnership, 253 

Equation  of  Payments, 257 

Alligation  Medial, 260 

Analysis, 261 

Exchange  of  Currencies, 269 

Duodecimals, 273 

Involution, 275 

Evolution, 277 

Extraction  of  the  Square  Eoot, 279 

Applications  of  the  Square  Eoot, 284 

Extraction  of  the  Cube  Root, 286 

Arithmetical  Progression, 293 

Geomf.trical  Progression, 296 

Permutation, 299 

Mensuration. 

Definitions, 300 

Measurement  of  Surfaces, 302 

Plasterers',  Painters',  and  Carpenters' Work, 803 

Measurement  of  Bodies  or  Solids, 306 

Masons*  and  Bricklayers'  "Work, 308 

Gauging, 31] 

Promiscuous  Questions, 314 


obseryatio:n'S  to  teachers. 

Intellectual  Arithmetic  should  be  thoroughly  studied  by  all, 
and  especially  by  the  young,  before  commencing  Practical.  For 
this  purpose,  attention  is  called  to  "  Ray's  Arithmetic,  Second 
Book,"  which  has  been  carefully  prepared,  and  is  now  published 
with  important  improvements. 

"When  admissible,  pupils  studying  Arithmetic  should  be  taught 
in  classes;  the  presence  of  the  class  being  a  stimulus  to  both 
teacher  and  pupil.  This  arrangement  also  economizes  time,  since 
the  same  oral  illustrations^  necessary  for  the  instruction  of  a  single 
pupil,  serve  for  a  class. 

The  time  occupied  at  each  recitation  ought  not  to  be  less  than 
thirty  minutes,  nor  more  than  one  hour.  The  class  should  not  be 
too  large;  and,  if  possible,  the  attainments  of  its  members  equal. 

Every  school  should  have  a  blackboard,  on  which  pupils  can 
Bolve  the  questions  and  explain  the  method  of  solution. 

A  prime  object  in  recitations  is  to  secure  attention;  to  do  this, 
the  exercises  must  be  interesting,  and  all  must  be  kept  employed. 
Let  as  many  be  called  out  as  can  obtain  positions  at  the  black- 
board, and  let  all  solve   the  same  question  at  once. 

When  the  solutions  are  completed,  let  some  one  be  called  on  to 
explain  the  process,  giving  the  reason  for  each  step  of  the  opera- 
tion. Exercises  thus  conducted  animate  the  class;  and  by  re- 
quiring the  learner  to  explain  every  process,  and  assign  a  reason 
for  every  step,  he  learns  to  rely  on  his  own  reasoning  powers. 

In  assisting  pupils  to  overcome  difficulties,  it  is  preferable  to 
do  it  indirectly,  by  making  such  suggestions,  or  asking  such 
questions,  as  will  enable  the  learner  to  accomplish  the  object. 

Frequent  Reviews  will  be  found  of  great  benefit. 

The  pupil  should  be  rendered  familiar  with  the  answers  to  the 
questions  in  the  Review  at  the  foot  of  the  page.  This  review  is 
intended  to  aid  the  teacher,  but  not  to  prevent  his  asking  other 
questions,  or  presenting  different  illustrations. 


ARITHMETIC. 

1.  A  Unit,  or  one,  is  a  single  thing  of  any  kind;  as, 
one  apple,  one  dollar,  one  pound. 

2.  Number  is  a  term  signifying  one  or  more  units  ; 
as,    one,    five,    seven  cents,   nine  men. 

3.  Numbers  are  expressed  by  ten  characters,  called 
Figures;     as,  1,  2,  3,  4,  5,  6,  7,  8,  9,  0. 

4.  Arithmetic  treats  of  numbers,  and  is  the  art  of 
computing  by  them.  The  fundamental  rules  are  five  ; 
Notation  and  Numeration,  Addition,  Subtraction,  Multi- 
plication, and  Division. 

I.  NOTATION  AND  NUMERATION. 

Art.  1.  A  single  thing  is  a  Unit,  or     o??e;  written,  1 
One       unit     and  one  more,   are     tv:o ;       ..       2 


Two  units  and  one  more,  are  three; 

Three  units  and  one  more,  are  four; 

Four  units  and  one  more,  are  five; 

Five  units  and  one  more,  are  six; 

Six  units  and  one  more,  are  seven; 

Seven  units  and  one  more,  are  eight; 

Eight  units  and  one  more,  are  nine ; 


3 
4 
5 

6 

7 
8 
9 


Art.  2.  These  nine  characters  are  called  digits^  or 
significant  figures^  because  they  denote  something. 

The  character,  0,  called  cipher,  naught,  or  zero,  is  em- 
ployed to  denote  nothing :  thus,  to  show  that  there  are 
no  cents  in  a  purse,  write,  the  niimher  of  cents  is  0. 

Remark. — The  cipher  is  sometimes  termed  an  auxiliary  digit, 
because  it  helps  the  other  digits  in  expressing  numbers. 

Review. — Art.  1.  "What  is  a  single  thing  ?  What  are  one  unit  and 
one  more  of  the  same  kind  ?  Two  units  and  one  more,  &c.  ?  2.  What 
are  these  nine  characters  called  ?  Why  ?  What  does  naught,  or  zero 
denote  ?    Rem.  What  is  the  cipher  termed  ?    Why  ? 

(9) 


10 


RAY'S   PRACTICAL   ARITHMETIC. 


it 


ORDER    OF    TENS. 

Art.  3.  Nine   units  and  one  more  are  called   Ten: 
forms  a  unit  of  a  second  or  higher  order  called  Tens. 

Ten  is  represented  by  the  same  figure,  1,  as  a  single 
thing,  or  unit  of  the  first  order ;     but. 

To  distinguish  ten,  the  1  is  written  in  the  second  order, 
the  second  place  from  the  right  hand  : 

The  first  order,  or  right  hand  place,  is  filled  with  a 
cipher.  The  cipher,  in  the  first  place,  denotes  that  there 
are  no  units  of  the  first  order. 

The  number  ten  is  written  thus ; 


unit,  are 


units, 
units, 
units, 
units, 
units. 


One  ten  and  one 

One  ten  and  two 

One  ten  and  three 

One  ten  and  four 

One  ten  and  five 

One  t«n  and  six 

One  ten  and  seven  units. 

One  ten  and  eight   units, 

One  ten  and  nine     units, 

Two  tens, 

Two  tens  and  one  unit,  . 

Two  tens  and  five  units, 

Three  tens. 

Four    tens, 

Five     tens. 

Six       tens. 

Seven  tens, 

Eight  tens. 

Nine    tens. 


...  10 

eleven ;  11 

twelve ;  12 

thirteen ;  13 

fonrteen ;  14 

fifteen ;  15 

sixteen ;  1  G 

seventeen ;  17 

eighteen ;  18 

nineteen ;  19 

twenty ;  20 

twenty-one  ;  2 1 

twenty-five  ;  2  5 


thirty  ; 
forty; 

sixty ; 
seventy ; 
eighty  ; 
ninety  ; 


30 
40 

50 
60 
70 
80 
90 


Review. — 3.  What  are  nine  units  and  one  more  called  ?  Ten  forma 
a  unit  of  what  order  ?  How  is  it  written  ?  When  two  figures  are  written 
together,  what  is  the  place  on  the  right  called  ?  Aiis.  The  first  place, 
or  first  order,  or  unit's  place.  What  the  place  on  the  left  ?  Ans.  The 
second  place,  or  second  order,  or  tens'  place. 

3.  In  writing  ten,  why  is  a  naught  put  in  unit's  place  ?  What  are  one 
ten  and  ono  unit  ?     liuw  written  i     What  are  two  tons  ?     How  written  ? 


NOTATION   AND  NUMERATION.  11 

Art.  4.  The  immbers  between  20  and  30,  30  and  40, 
&c.,  may  be  expressed  by  considering  the  tens  and  units 
of  which  they  are  composed. 

Thus,  the  number  twenty-six  is  composed  of  two  tens 
and  6  units ;  and  is  expressed  by  writing  2  in  the  place 
of  tens,  and  6  in  the  place  of  units,  thus, 2  6 

Ninety-seven  has  9  tens  and  7  units  ;  written,        9  7 

ORDER    OF    HUNDREDS. 

Art,  5.  Ten  tens  are  One  Hundred,  which  forms  a  unit 
of  the  third  order  called  Hundreds. 

It  is  represented  by  the  same  figure,  1,  as  a  single 
thing,  or  unit  of  the  first  order;     but. 

The  1  is  written  in  the  third  order,  while  the  orders  of 
tens  and  units  are  each  filled  with  a  cipher,  thus,  10  0. 

The  mode  of  writing  one  unit,  two  units ;  one  ten,  two 
tens,  &c.,  has  been  explained. 

In  writing  one  hundred,  two  hundreds,  &c.,  place  the 
figure  representing  the  hundreds  in  the  third  order,  and 
fill  the  place  of  units  and  tens  with  ciphers,  thus : 


Two  hundred  ....  2  0  0 
Three  hundred  ...  3  0  0 
Four  hundred  ...  4  0  0 
Five  hundred.  ...500 


Six  hundred  ....  6  0  0 
Seven  hundred  ...  7  0  0 
Eight  hundred  ...  8  0  0 
Nine  hundred    ...  9  0  0 


Art.  6.  "With  the  three  orders  of  units,  tens,  and 
HUNDREDS,  all  the  numbers  between  one  and  one  thou- 
sand may  be  expressed. 

In  the  number  three  hundred  and  sixty-five,  are  three 
hundreds,  six  tens  or  sixty,  and  5  units ;  written,  3  6  5. 

The  number  four  hundred  and  seven,  contains  4  hun- 
dreds, 0  tens,  and  7  units;  written,  4  07. 

In  the  number  seven  hundred,  are  7  units  of  the  third 
order,  and  no  units  of  the  first  and  second  orders,  7  00. 

Eeview. — 4.  How  arc  the  numbers  between  20  and  .30,  30  and  40, 
written  ?  5.  Of  what  order  do  ten  tens  form  a  unit  ?  In  writing  one 
hundred,  what  orders  are  filled  with  ciphers  ? 


12  RAY'S    PRACTICAL    ARITHMETIC. 

ORDER    OF    THOUSANDS. 

Art.  7.  Ten  hundreds,  or  ten  units  of  the  order  of 
hundreds,  are  called  One  Thousand^  which  forms  a  unit 
of  the  fourth  order. 

One  thousand  is  expressed  thus, 100  0 

Two  thousand,  thus, 2000 

Three  thousand,  thus, 3000 

Art.  8.  Principles  of  Notation  and  Numeration. 

1.  Numbers  are  represented  by  the  9  digits  and  cipher. 

2.  The  cipher  or  naught  (0)  has  no  value ;  it  is  used 
merely  to  fill  vacant  orders. 

3.  The  number  expressed  by  any  figure  depends  upon 
the  order  or  place  which  it  occupies ; 

Thus,  2  in  the  first  order  is  2  units  ;  in  the  second  order, 
2  tens  or  twenty;  in  the  third  order,  2  hundreds,  &c. 

4.  The  number  expressed  by  a  figure  standing  alone,  is 
called  its  simple  value  ;  the  number  expressed  when  com- 
bined with  other  figures,  its  local  value. 

5.  The  local  value  of  a  figure  increases  from  right  to 
left  tenfold:  ten  units  of  the  first  order  make  one  unit  of 
the  second;  ten  units  of  the  second,  one  of  the  third. 

Invariably,  ten  units  of  any  order  make  one  unit  of  tJie 
next  higher  order. 

NoTK. — When  units  are  named,  those  of  the  first  order  are  meant. 

Art.  9.  The  name  of  each  of  the  first  nine  orders  is 
learned  from  the  Table  of  Orders. 

Review. — 6.  With  what  three  orders  may  the  numbers  between  one 
and  one  thousand  be  expressed  ?  In  writing  the  number  three  hundred 
and  five,  what  order  would  be  filled  with  a  cipher  ?     Why  ? 

7.  Ten  hundreds  make  a  unit  of  what  order  ?  H' w  expressed  ?  In 
writing  one  thousand,  what  orders  are  filled  with  ciphers  ?     Why  ? 

8.  By  what  characters  are  numbers  represented  ?  What  value  has  the 
cipher  ?  Fur  what  is  it  used  ?  Upon  what  does  the  number  expressed  by 
a  figure  depend  i  What  is  2  in  the  first  pla<^e  ?  In  the  second  ?  In  the 
third  ?  What  is  the  .simple  value  of  a  figure  ?  The  local  value  ?  How 
does  it  increase  ?     What  is  meant  by  tenf»>ld  ? 

9.  Where  learn  the  names  of  the  first  nine  orders?     Repeat  the  table. 


NOTATION    AND  NUiMERATION. 


13 


TABLE    OF    ORDERS. 
9tli.     8tli.     7tli.     6th.     5th.     -Ith.     3d.      2d.     1st. 


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—1 

4 

^ 

k<H 

rt 

H 

:3 

»                  . 

<+- 

o 

o 

A 

cc 

■73 

H 

c 

XT. 

rH 

•73 

o 

C4_ 

CJ 

':3 

o 

* 

OQ 

QQ 

r-i 

cn 

-M 

a 

c 

<— 

Art.  10.  For  convenience,  the  different  orders  are 
divided  into  ^periods^  of  three  orders  each. 

The  first  three  orders,  that  is,  Units,  Tens,  Hundreds, 
constitute  the  first  or  Unit  Period. 

The  second  three  orders,  that  is,  Thousands,  Tens  of 
Thousands,  Hundreds  of  Thousands,  constitute  the  second 
or  Thousand  Period. 

The  next  three  orders,  that  is.  Millions,  Tens  of  Mil- 
lions, and  Hundreds  of  Millions  constitute  the  third  or 
Million  Period. 


Art.  11.  Periods  according  to  the  Common  Method  : 


First      period,  Unit. 
Second  period,  Thousand. 
Third     period,  Million. 
Fourth  period,  Billion. 
Fifth      period,  Trillion. 
Sixth      period,  Quadrillion. 


Seventh    period,  Quintillion. 
Eighth      period,  Sextillion. 
Ninth       period,  Septillion. 
Tenth        period,  Octillion. 
Eleventh  period,  Nonillion. 
Twelfth     period,  Decillion. 


For  other  methods  of  Numeration,  see  "  Ray's  Higher  Arithmetic. ^^ 

Review. — 10.  Why  are  the  orders  divided  into  periods  ?  How  many 
orders  constitute  a  period  ?  "What  is  the  name  of  the  first  period,  and  of 
what  orders  is  it  composed  ?  The  name  of  the  second  period,  and  of  what 
orders  is  it  composed  ?    The  name  of  the  third  period  ? 


14 


RAY'S  PRACTICAL   ARITHMETIC. 


Art.  12.  This  Table  shows  the  division  into  Periods. 


5th  Period. 

Trillions. 


4th  Period. 
Billions. 


3d  Period. 
Millions. 


2d  Period. 
Thousands. 


1st  Period. 
Units. 


c 
o 

^   c 
^  .2 


«  ^     C 

K  H  H 
6  5  4 


00 

C 
o 

«   5 


tH  o  a 

^  a  ^ 

K  H  S5 

3  2  1 


C 

.2 

'^    o 

<«  ::3 
o  rs 

^   o  5 

=3      0   3 

9  8  7 


c  • 

c3  . 

3  00 

O  t3 

^  a 

o  2 


30 


00 

a 

CO 

3 
O 

m  H  H 


00  ^ 


CO 


m 


6  5  4 


00 

T3 

V 

3   5j  a 
3  2  1 


Art.  13.  The  art  of  reading  numbers  when  they  are 
written  in  figures,  is  called 

NUMERATION. 

Rule  for  Numeration. — 1.  Begin  at  the  right  hand,  and 
point  off  the   numbers  into  periods   of  three  Jigures  each. 

Beginning  at  the  left  hand,  read  each  period  as  if  it  stood 
alone  ;  then  add  the  name  of  the  period. 

10,  ten;  or  one  ten  and  no  units. 

12,  twelve;  or  one  ten  and  two  units. 

25,  twenty-five;  or  two  tens  and.  five  units. 

6  0,  sixty;  or  six  tens  and  no  units. 

200,  tico  hundred;  or  2  hundreds,  0  tens,  and  0  units. 

305,  three  hundred  and  five;  or  3  hundreds,  0  tens, 
and  5  units. 

7405,  seven  thousand  four  hundred  and  five  ;  or  7  thou- 
sands, 4  hundreds,  0  tens,  and  5  units. 

45  0  0  8,  forty-five  thousand  and  sixty-eight;  or  4  tens  of 
thousands,  5  thousands,  0  hundreds,  6  tens,  and  8  units. 


KKVIRW.---12.  Namo  tho  periods.     13.  What  is  Numeration  ?    Rule? 


NOTATION    AND  NUMERATION. 


15 


EXAMPLE. 


368271927 


Read  thus, 


i 


Q 

3 


Si 


I      -    C    '^ 

3     6 


o 


a 

.HP 

8 


o 
(-1 

-«  H 

OT3  ^ 

^    C  n, 

-*-    «3  OQ 


a 

w 
O 

o 


9 


a 

Ol 
CD 


2     7      1       9     2     7 


EXAMPLES  IN  NUMEEATION, 
To  he  copied^  separated  into  periods y  and 


read. 


5 

901 

10000 

200000 

20020 

300300 

63 

1000 

12000 

456000 

106307 

909090 

90 

1005 

13200 

682300 

400001 

8600050 

100 

1050 

50004 

704208 

302404 

2102102 

104 

1085 

62001 

800141 

800010 

4080400 

147 

1100 

70400 

900016 

700010 

1000011 

208 

1108 

80090 

601020 

1030725 

3036000 

280 

3003 

97010 

700400 

4050607 

9001003 

403 

4050 

40305 

800002 

6601000 

33007820 

729 

3045 

76052 

910103 

7406035 

61189602 

710 

9699 

83991 

700100 

9725014 

99099099 

130670921 

23004090901 

6900702009 

942029307029 

Art.  14.  Examples  to  he  written  in  figures^  then  read. 

1.  One  unit  of  the  third  order. 

2.  Two  units  of  the  third  order,  and  three  units  of  the 
first  order. 

3.  Three  units  of  the  fourth  order,  and  two  units  of 
the  second  order. 

4.  Five  units  of  the  fifth  order,  two  units  of  the  third 
order,  and  four  units  of  the  first  order. 

5.  Nine  units  of  the  seventh  order,  and  five  units  of  the 
third  order. 

6.  Seven  units  of  the  fifth  order,  and  five  units  of  the 
second  order. 


16 


RAY'S   PRACTICAL   ARITHMETIC. 


7.  Six  units  of  the  ninth  order,  and  four  units  of  the 
fourth  order. 

8.  Two  units  of  the  eighth  order,  five  units  of  the  fifth, 
and  two  units  of  the  the  third. 


9.   What  is  5  in  the  first  order? 
3  in  the  sixth  ?  8  in  the  third  ? 

9  in  the  seventh  ?        4  in  the  fifth  ? 
8  in  the  sixth?  4  in  the  ninth? 


4  in  the  second  r 
7  in  the  fourth  ? 
2  in  the  eighth  1 

5  in  the  tenth  ? 


Art.  15.  The  art  of  expressing  numbers  by  figures  or 
letters,  is  called 

NOTATION. 

Rule  for  Notation. —  Wj-ite  each  figure  in    the  order  to 
which  it  belongs^  and  Jill  the  vacant  orders  with  ciphers. 

Example. — Write  in  figures,    the    number    forty-five 
thousand  and  twenty-six. 

Put  4  in  ten  thousands'  place  for  forty  thousand ; 
5  in  thousands'  place  for  five  thousand ; 
0  in  hundreds'  place,  as  there  are  no  hundreds ; 
2  in  tens'  place  for  twenty ;  and, 


6  in  units'  place  for  six 


written,  45  026. 


EXAMPLES  IN  NOTATION. 

Note. — Learners  sliould  be  taught  to  put  a  dot  for  each  order, 
and  a  large  dot  for  the  commencement  of  each  period. 


1.  Four  units. 

2.  One  ten  and  six  units. 

3.  ?]ighteen. 

4.  Two  tens,  or  twenty. 

5.  Two  tens  and  four  units, 
fi.  Twenty-eight. 

7.  Three  tens,  or  thirty. 

8.  Three  tens  and  two  units, 
y.  Thirty-seven. 

10.  Four  tens  and  one  unit. 

11.  Forty-six. 

12.  Five  tens  and  nine  units. 


13.  Sixty-five. 

14.  Nine  tens  and  seven  units. 

15.  Eighty-seven. 

16.  One  hundred  and  four. 

17.  One  hundred  and  two  tens. 

18.  One  hundred  and  thirty. 

19.  One  hundred  and  seventy- 
five. 

20.  Two    hundreds  and  three 
units. 

21.  Three  hundreds  and  four 
tens. 


NOTATION   AND  NUMERATION 


17 


22.  Four  hundreds,  three  tens, 
and  five  units. 

23.  Five  hundred  and  two. 

24.  Six  hundred  and  twenty- 
five. 

25.  Two  hundreds,  two  tens, 
and  two  units. 

26.  Nine  hundreds,  nine  tens, 
and  nine  units. 

27.  Eio;ht  hundred  and  seven. 

28.  Eight  hundred  and  seventy. 

29.  Nine  hundred  and  one. 

30.  Six  hundreds  and  six  units. 

31.  Three  hundred  and  nine. 

32.  One  hundred  and  ninety. 

33.  Two  hundred  and  two. 

34.  One  unit  of  the  third  order 
and  one  unit  of  the  first 
order. 

35.  Two  units  of  the  second 
order  and  one  unit  of  the 
first  order. 

36.  Three  units  of  the  third 
order.  What  orders  have 
ciphers  in  them  ? 

37.  Five  units  of  the  third 
order  and  three  units  of 
the  first  order. 

38.  Eight  units  of  the  third 
order,  two  units  of  the 
second  order,  and  seven 
units  of  the  first  order. 

39.  One  unit  of  the  fourth 
order. 

40.  One  thousand  and  twenty 
— or  one  thousand,  no  hun- 
dreds, two  tens,  and  no 
units. 

41.  Twentv-five  thousand  and 
SIX — or  two  tens  of  thou- 
sands, five  thousands,  no 
hundreds,  no  tens,  and  six 
units. 

3d  Ek.  2 


42.  Three  hundred  and  forty- 
five —  or  three  hundreds, 
four  tens,  and  five  units. 

43.  Seven  hundred  and  sixty 
— or  seven  hundreds,  six 
tens,  and  no  units. 

44.  Three  thousand  four  hun- 
dred and  six — or  three 
thousands,  four  hundreds, 
no  tens,  and  six  units. 

45.  Forty-two  thousand  and 
thirt}" — or  four  tens  of 
thousands,  two  thousands, 
no  hundreds,  three  tens, 
and  no  units. 

46.  Thirty  thousand — or  three 
tens  of  thousands,  no  hun- 
dreds, no  tens,  no  units. 

47.  One  hundred  and  sixty- 
three  thousand — or  one 
hundred  thousand,  six  tens 
of  thousands,  three  thou- 
sands, no  hundreds,  no 
tens,  and  no  units. 

48.  Three  hundred  and  forty- 
one  thousand,  five  hundred 
and  sixty-three — or  three 
hundred  thousands,  four 
tens  of  thousands,  one 
thousand,  five  hundreds, 
six  tens,  and  three  units. 

49.  Four  tens  of  millions,  five 
millions,  no  hundreds  of 
thousands,  eight  tens  of 
thousands, three  thousands, 
no  hundreds,  two  tens,  and 
six  units. 

50.  Eight  hundreds  of  millions, 
seven  tens  of  millions,  no 
millions,  seven  hundreds 
of  thousands,  four  tens  of 
thousands,three  thousands, 
five  hundreds,  seven  tens, 
and  nine  units. 


18 


RAYS   PRACTICAL   ARITHMETIC. 


51.  Two  thousand  eight  hun- 
dred and  four. 

52.  Four  thousand  and  twenty- 
nine. 

53.  Six  thousand  and  six. 

54.  Twenty-two  thousand  seven 
hundred  and  sixty-five. 

55.  Eighty  thousand,  two  hun- 
dred and  one. 

56.  Ninety  thousand  and  one. 

57.  Thirty  thousand  and  thirty. 

58.  Four  hundred  and  ten 
thousand,  two  hundred  and 
five. 

59.  Eight  hundred  thousand, 
six  hundred  and  sixty-nine. 

60.  Nine  hundred  thousand 
and  one. 

61.  Five  hundred  thousand  and 
fifty. 


62.  One  hundred  thousand  and 
ten. 

63.  Nine  hundred  and  nine 
million  and  ninety  thou- 
sand. 

64.  One  hundred  million,  ten 
thousand  and  one. 

65.  Ninety-one  million,  seven 
thousand  and  sixty. 

66.  Seventy  million  and  four. 

67.  Seven  hundred  millions, 
ten  thousand  and  one. 

68.  One  billion,  one  million 
and  forty. 

69.  Forty  billions,  two  hundred 
thousand  and  five. 

70.  Seven  hundred  and  twenty- 
six  billions,  fifty  millions, 
one  thousand,  two  hundred 
and  forty-three. 


EOMAN    NOTATION. 

Art.  16.  The  common  method  of  representing  num- 
bers, by  figures^  is  termed  the  Arabic.  Another  method, 
by  means  of  letters^  is  termed  the  Roman. 

The  letter  I  stands  for  one;  Y,Jive;  X,  ien ;  L,  fifty; 
C,  one  hundred;  D,  five  hundred;  and  M,  one  thousand. 
Numbers  are  represented  on  the  following  principles  : 

1.  Every  time  a  letter  is  repeated,  its  value  is  repeated  : 

thus,    II  denotes  two,    XX  denotes  twenty. 

2.  When  a  letter  of  less  value  is  placed  before  one  of 
greater  value,  the  less  is  taken  from  the  greater  ;  if 
placed    after   it,    the  value  of  the  greater   is   increased  : 

thus,    IV  denotes  four,    while  VI  denotes  six. 

Rkvikw.— 15.  What  is  Notation?  What  is  the  Rule?  16.  What  is 
the  common  method  of  Notation  termed  ?  What  other  method  ?  What 
number  is  represented  by  the  letter  I?  By  V  ?  By  X  ?  By  L  ?  By  C  ? 
By  D  ?     By  M  ?     What  is  the  effect  of  repeating  a  letter  ? 

Id.  What  the  effect  of  placing  a  letter  of  less  value  before  one  of  greater 
value  ?    After  one  of  greater  value  ?    What  of  placing  a  lino  over  a  letter  ? 


ADDITION    OF    SLMPLE   NUMBERS. 


19 


3.  A  line  over  a  letter  increases  its  value  a  thousand 
times.     Thus,  V  denotes  5  0  0  0,  M  denotes  07ie  million. 

TABLE    OF    EOMAN    NOTATION. 


I     .  . 

,    .     One. 

XXI    . 

Twenty-one. 

II    .  . 

.     Two. 

XXX  . 

.      Thirty. 

Ill  .   . 

,     Three. 

XL. 

Forty. 

IV     .    . 

.     Four. 

L     .     . 

Fifty. 

V      .    .    . 

•     ?.^^'^- 

LX.     . 

.     Sixty. 

VI     .     . 

.     Six. 

XC.     . 

Ninety. 

IX     . 

,     .     Nine. 

C     .     .     . 

One  hundred. 

X      .     . 

.     Ten. 

CCCC.     . 

Four  hundred. 

XI     . 

.     Eleven. 

D         .     . 

Five  hundred. 

XIV  .     .    . 

.     Fourteen. 

DC.     . 

Six  hundred. 

XV    .     . 

.     Fifteen. 

DCC    .     . 

Seven  hundred 

XVI  .     . 

.     .     Sixteen 

DCCC      . 

Eight  hundred 

XVII     . 

.     Seventeen. 

DCCCC 

Nine  hundred. 

XVIIl    .    . 

,     Eighteen. 

M    .     . 

One  thousand. 

XIX.     .     . 

,     Nineteen. 

MM     .    . 

Two  thousand. 

XX   .     .    . 

.     Twenty. 

MDCCCL 

VI      .  1856. 

Art.  17.   The   preceding  illustrations  show  the  three 
methods  of  expressing  numbers  : 

1st.  By  words,  or  ordinary  language. 

2d.     'By  figures^  termed  the  Arabic  method. 

3d.    By  letters^   termed  the  Roman  method. 


II.   ADDITION. 

1.  If  you  have  2  cents  and  find  3  cents,  how  many 
will  you  then  have  ?  Ans.  5  cents. 

2.  I  spent  12  cents  for  a  slate,  and  5  cents  for  a  copy- 
book :  how  many  cents  did  I  spend  ?         Ans.  17  cents. 

3.  John  gave  6  cents  for  an  orange,  7  cents  for 
pencils,  and  9  cents  for  a  ball :  how  many  cents  did  they 
all  cost  ?  Ans.  22  cents. 

Art.  18.  The  process  of  uniting  two  or  more  numbers  into 
one  number,  is  termed  Addition. 

The  number  obtained  by  addition,  is  the  Sum  or  Amount. 


^ 


20 


RAY'S    PRACTICAL    ARITHMETIC. 


Remark. — When  the  numbers  to  be  added  are  of  the  same  de- 
nomination, that  is,  all  cents,  or  all  yards,  &c.,  the  operation  is 
called  Simple  Addition. 

Art.  19.     OF    THE    SIGITS. 

The  Sign  +,  called  plus^  means  more.  When  between 
two  numbers,  it  shows  that  they  are  to  be  added ;  thus, 
4  +  2  means  that  4  and  2  are  to  be  added  together. 

The  sign  of  equality^  =,  denotes  that  the  quantities 
between  which  it  stands  equal  each  other. 

The  expression  4+2=6,  means  that  the  sum  of  4  and 
2  is  6  ;     read,  4  and  2  are  6,  or  4  plus  2  equals  6. 

ADDITION    TABLE. 


2  +  0=    2 

3  +  0-    3 

4  +  0=    4 

5  +  0=    5 

2  +  1-    3 

3  +  1-    4 

4  +  1=5 

5  +  1=6 

2  +  2=    4 

3  +  2-    5 

4+2-    6 

5  +  2=    7 

2  +  3=    5 

3  +  3        6 

4  +  3—    7 

5  +  3=    8 

2  -h4=    6 

3  +  4-    7 

4  +  4=    8 

5  +  4=    9 

2+5=7 

3  +  5=    8 

4  +  5=    9 

5  +  5  =  10 

2  f  6  =   8 

3  +  6-    9 

4  +  6  =  10 

5  +  6  =  11 

2  h7-    9 

3  +  7-10 

4  +  7  =  11 

5  +  7  =  12 

2  +  8  =  10 

3  +  8-11 

4  +  8  =  12 

5  +  8  =  13 

2  +  9  =  11 

3+9-12 

4  +  9  =  13 

5  +  9  =  14 

6  +  0=    6 

7  +  0=    7 

8+0=   8 

9  +  0=    9 

6  +  1-7 

7  +  1=    8 

8+1=    9 

9  +  1  =  10 

6  +  2=    8 

7  +  2=    9 

8  +  2  =  10 

9  +  2  =  11 

6  +  3—    9 

7  +  3  =  10 

8  +  3  =  11 

9  +  3  =  12 

6  +  4-10 

7+4-11 

8  +  4  =  12 

9  +  4  =  13 

6  +  5  =  11 

7  +  5  =  12 

8  +  5  =13 

9  +  5-14 

6  +  6-12 

7  +  6-13 

8  +  6  —  14 

9  +  6-15 

6  +  7-13 

7  +  7  =  14 

8  +  7-15 

9+7-16 

6  +  8  =  14 

7  +  8-15 

8  +  8  —  16 

9  +  8=  17 

6  +  9  =  15 

7  +  9  =  16 

8  +  9-17 

9  +  9-18 

Rkview. — 17.  What  aro  th»  throo  methods  of  expressing  numbers? 
18.  What  is  Additiun  ?  Wha  the  sum  or  amount?  Rem.  What  is 
Simple  Addition?  19,  What  does  the  sign  plus  mean?  What  does  it 
show?     Wlmt  does  the  sign  of  equality  denote?     (Jive  an  example. 


ADDITION    OF    SIMPLE    NUMBERS.  21 

Art.  20.  1.  James  had  63  cents,  and  his  father  gave 
him  35  cents  :  how  many  cents  had  he  then  ? 

Place  the  units  and  the  tens  of  one  number  under  the 
units  and  the  tens  of  the  other,  that  figures  of  the  same 
unit  value  may  be  more  easily  added. 

Solution. — Write   the   numbers  as  in    the  g  .| 

margin ;  then  say  5  units  and  3  units  are  8  -^  =* 

units,  which  write  in  units'  place;  3  tens  and  cents. 

6  tens  are  9  tens,  which  write  in  tens'  place.  ^_  ^^^^^s- 

The  sum  is  9  tens  and  8  units,  or  98  cents.  Ans.  9  8  cents. 

In  this  example,  7mits  are  added  to  units,  and  tens  to  tens, 
since  only  numbers  of  the  same  kind,  that  is,  having  the 
same  unit  value,  can  be  added.     Thus, 

3  units  and  2  tens  make  neither  5  units  nor  5  tens ;  as, 
3  apples  and  2  plums  are  neither  5  apples  nor  5  plums. 

2.  I  own  3  tracts  of  land  :  the  first  contains  240  acres  ; 
the  second,  132  acres ;  the  third,  25  acres  :  how  many 
acres  in  all  ? 

Since  units  of  different  orders  can  not  be  added  to- 
gether, place  units  under  units,  tens  under  tens,  &c.,  that 
figures  to  be  added  may  be  in  the  most  convenient  position. 

Solution. — Begin  at  the  right,  and  say  5  units  2  4  0  acres, 
and  2  units  are  7  units,  which  write  in  units'  13  2  acres, 
place;  2  tens  and  3  tens  and  4  tens  are  9  tens,  2  5  acres, 

which  write  in  tens'  place  : 

Lastly,  1  hundred  and  2  hundreds  are  3  hun-  ^^'^  *^^^^®- 
dreds,  which  write  in  hundreds'  place,  and  the  work  is  complete. 

Questions  to  be  solved  and.  explained  as  above. 

The  sign  $  stands  for  the  word  dollars,  and  when  used 
is  placed  before  the  figures. 

3.  There  are  43  sheep  in  one  pasture,  21  in  another, 
and  14  in  another  :  how  many  sheep  in  all  ?      Ans.  78. 

Review. — 20.  Why  are  units  placed  under  units,  and  tens  under  tens  ? 
Can  numbers  of  different  unit  values  be  added  ?  Why  not?  Give  an  ex- 
ample.     What  sign  is  used  for  the  word  dollars  ? 


22  RAY'S   PRACTICAL    ARITHMETIC. 

4.  I  owe  one  man  $210,  another  $142,  and  another 
835  :  what  is  the  amount  of  the  debts  ?  Ans.  $387. 

5.  Find  the  sum  of  4321,    1254,    3120.     Ans.  8695. 

6.  The  sum  of  50230,   3105,   423.  Ans.  53758. 

Art.  21.  Where  the  sum  of  the  figures  in  a  column 
does  not  exceed  9,  it  is  written  under  the  column  added. 

When  the  sum  of  the  figures  exceeds  9,  two  or  more 
figures  are  required  to  express  it.    To  explain  the  method, 

TAKE   THIS   EXAMPLE. 

1.  James  bought  a  reader  for  74  cents,  an  atlas  for  37 
cents,  a  slate  for  25  cents  :  how  much  did  all  cost  ? 

1st  Solution. — By  adding  the  figures  in 
the  first  column,  the  sum  is  16,  which  is  1 
ten  and  6  units.  Write  the  6  units  in  the 
order  of  units,  and  the  1  ten  in  the  order  of 
tens. 

The  sum  of  the  figures  in  the  tens'  column 
is  12  tens,  which  is  1  hundred  and  2  tens. 
Write  the  2  tens  in  the  order  of  tens,  and       ^"*-  1^6  cents, 
the  1  hundred  in  the  order  of  hundreds. 

Lastly,  unite  the  figures  in  the  column  of 
tens.     The  sura  is  1  hundred,  3  tens,  and  6  units,  or  136  cents, 

2d  Solution. — The  preceding  example  is  usually  per-  7  4 
formed  thus :  5  units  7  units  and  4  units  are  16  units,  3  7 
which  is  1  ten  and  6  units.  Write  the  6  units  in  units'  2  5 
place,  and  carry  the  1  ten  to  tens'  place.  

Then,  1  ten  2  tens  3  tens  and  7  tens  are  13  tens,  which      -^  ^  " 
is  1  hundred  and    3  tens;   write  the  3  tens  in   tens'  place, 
the  1  hundred  in  hundreds'  place,  and  the  work  is  completed. 

The  1  ten  derived  from  the  sum  of  the  fi2:ures  in  the 
1st  column,  and  added  to  the  2d,  is  said  to  be  carried. 

Review. — 21.  When  tho  sum  of  a  column  does  not  exceed  9,  where  is 
it  written?  If  greater  than  9?  What  is  understood  by  carrying  the 
tens?  In  what  does  it  consi.st?  Why  does  tho  addition  begin  with  the 
units'  column  ?     22.  What  is  tho  rule  for  addition  ?     The  proof* 


T.  V. 

Reader  7  4 
Atlas     3  7 
Slate      2  5 

cents, 
cents, 
cents. 

16 
12 

ADDITION    OF   SIMPLE   NUMBERS.  23 

When  the  sum  of  the  figures  in  a  column  exceeds  9, 
reservino;  the  tens,  or  left  hand  figure,  and  adding;  it  to 
the  figures  in  the  next  column,  is  called  carrying  the  tens. 

2.  Add  the  numbers,  3-il5,  503,  1870,  and  922. 

Solution. — Write  units  of  the  same  order  under  3415 
each  other.  Then  say,  2  and  3  are  5,  and  5  are  10  5  0  3 
units,  which  is  no  (0)  units,  written  in  units'  place,  1870 
and  1  ten,  carried  to  the  tens;  1  and  2  are  3,  and  7        92  2 

are  10,  and  1   are  11   tens,  which  is  1  ten,  written  in     

tens'  place,  and  1  hundred,  carried  to  the  hundreds;      «  ^  lo 
1  and  9  are  10,  and  8  are  18,  and  5  are  23,  and  4  are  27  hundreds, 
which  is  7  hundreds,  written  in  hundreds'  place,  and  2  thousands, 
carried  to  the  thousands ;  2  and  1  are  3,  and  3  are  6  thousands, 
written  in  thousands'  place. 

Carrying  the  tens  is  simply  adding  (ens  to  tens,  hundreds 
to  hundreds,  &c.,  on  the  principle  (Art.  20),  that  only 
numbers  of  the  same  unit  value  can  be  added. 

The  addition  begins  at  the  right  hand  column,  with  the 
unit  of  the  lowest  order,  so  that, 

If  the  sum  of  the  units  in  any  column  exceeds  9,  the 
tens  can  be  carried  to  the  sum  of  the  next  hio-her  order. 


o' 


Art.  22.  eule 

For  Addition. — 1-  Write  the  numbers  to  he  added,  so  thai 
figures  of  the  same  order  may  stand  under  each  other,  units 
under  units,  tens  under  tens,  &c. 

2.  Begin  at  the  right  hand,  and  add  each  column  sepa- 
rately. Place  the  units  obtained  by  adding  each  column  under 
it,  and  carry  the  tens  to  the  next  higher  order.  At  the  last 
column  write  the  whole  amount. 

Proof. — Separate  the  numbers  into  two  or  more  divisions ; 
find  the  sum  of  each;  then  add  the  amounts  together.  If  their 
sum  equal  that  found  by  the  rule,  the  work  is  correct 

Remarks. — 1.  Another  method  of  proof  consists  in  adding  the 
columns  downward,  taking  the  figures  in  a  different  order  from 
that  in  which  they  were  taken  before. 


24 


RAYS    PRACTICAL   ARITHMETIC. 


2.  For  the  method  of  proof  by  casting  out  the  9's,  (too  difiScult 
for  a  work  of  this  grade,)  see  "  Ray  s  Higher  Arithmetic^ 

3.  Add  the  numbers  853,    516,    29,    6,   34,   580. 


Solution. — Writing  the  numbers  and  add- 
ing them,  their  sum  is  2018. 

Proof. — If  the  numbers  be  separated  by 
a  line  between  20  and  6,  the  sum  of  the  num- 
bers in  the  first  division  will  be  1398,  and 
in  the  second,  620.  Adding  these  numbers 
together,  their  sum  is  2018,  as  before. 


853 

516 

29 
g— 1398 

34       620 

580    2018 


2018 


4.  Find  the  sum  of  3745,    2831,    5983,  7665. 

In  adding  long  columns  of  figures,  it  3745 

is  necessary  to  retain  the  numbers  carried.  2831 

This   may  be  done   by  placing  them   in  5983 

small  figures  under  their  proper  columns,  7665 


J,  2,  1,  in 

the  margin. 

20224 

3  2  1 

EXAMPLES  FOR  PRACTICE. 

(5) 

(6) 

(7) 

(8) 

(9) 

(10) 

184 

204 

103 

495 

384 

1065 

216 

302 

405 

207 

438 

6317 

135 

401 

764 

185 

348 

5183 

320 

311 

573 

825 

843 

7102 

413 

109 

127 

403 

483 

3251 

101 

43 

205 

325 

834 
3330 

6044 

1369 

1370 

2177 

2440 

28962 

(11) 

(12) 

(13) 

(14) 

(15) 

3725 

5943 

82703 

95 

^7462 

6840325 

5834 

6427 

102 

47 

8345 

7314268 

4261 

8204 

6005 

610628 

3751954 

7203 

7336 

759 

423158 

6287539 

ADDITION    OF    SIMPLE   NUMBERS.  25 

16.  23+41-t- 7-4+83+16=  how  many?  Aiis.  237. 

17.  45+19+32+74+55=  how  many?  Ans.  225. 

18.  51+48+76+85+4  =  how  many?  Ans.  264. 

19.  263+104+321  +  155  =  how  many?  Ans.  843. 

20.  94753+2847+93688+9386+258+3456  are  how 
many  ?  Ans.  204388. 

21.  January  has  31  days,  February  28,  March  31, 
April  30,  and  May  31  :  how  many  days  are  there  in  these 
five  months?  Ans.  151. 

22.  June  has  30  days,  July  31,  August  31,  September 
30,  October  31 :  how  many  days  in  all?  Ans.  153. 

23.  The  first  5  months  have  151  davs,  the  next  5  have 
153  days,  November  has  30,  and  December  31  :  how  many 
days  in  the  whole  year  ?  Ans.  365. 

24.  I  bought  4  pieces  of  muslin  :  the  first  contained  50 
yards,  the  second  65,  the  third  42,  the  fourth  89  :  how 
many  yards  in  all  ?  Ans.  246  yds. 

25.  I  owe  one  man  8245,  another  $325,  a  third  8187,  a 
fourth  896  :  how  much  do  I  owe  ?  Ans.  8853. 

26.  General  Washington  was  born  A.  D.  1732,  and 
lived  67  years  :  in  what  year  did  he  die?        Ans.  1799. 

27.  From  the  creation  of  the  world  to  the  flood  was 
1656  years  ;  thence  to  the  siege  of  Troy,  1164  years  ; 
thence  to  the  building  of  Solomon*s  Temple,  180  years  ; 
thence  to  the  birth  of  Christ,  1004  years  :  in  what  year 
of  the  world  did  the  Christian  era  begin  ?       Ans.  4004. 

28.  A  has  4  flocks  of  sheep ;  in  the  1st  are  65  sheep 
and  43  lambs;  in  the  2d,  187  sheep  and  105  lambs;  in 
the  3d,  370  sheep  and  243  lambs ;  in  the  4th,  416  sheep 
and  95  lambs  :  how  many  sheep  and  lambs  has  he? 

Ans.  1038  sheep,  and  486  lambs. 

29.  A  man  bought  30  barrels  of  pork  for  8285,  18 
barrels  for  8144,  23  barrels  for  8235,  and  34  barrels  for 
8408  :  how  many  barrels  did  he  buy,  and  how  many  dol- 
lars did  he  pay?  Ans.  105  barl.,  and  81072. 


26  RAY'S    PRACTICAL   ARITHMETIC. 

30.  The  first  of  four  numbers  is  287  ;  the  second,  596  ; 
the  third,  841 ;  and  the  fourth,  as  much  as  the  first  three  : 
what  is  their  sum  ?  Ans.  3448. 

31.  The  Pyramids  of  Egypt  were  built  337  years  before 
the  founding  of  Carthage ;  Carthage  was  founded  49 
years  before  the  destruction  of  Troy ;  Troy  was  destroyed 
431  years  before  Rome  was  founded ;  Carthage  was  de- 
stroyed 607  years  after  the  founding  of  Rome,  and  146 
beibre  the  Christian  era.  How  many  years  before  Christ 
were  the  Pyramids  built?  Ans.  1570. 

32.  Add  three  thousand  and  five ;  forty-two  thousand, 
six  hundred  and  twenty-seven  ;  105 ;  three  hundred  and 
seven  thousand  and  four ;  800,791 ;  three  hundred  and 
twenty  thousand,  six  hundred.  Ans.  1474132. 

33.  Add  275,432  ;  four  hundred  and  two  thousand 
and  thirty ;  three  hundred  thousand  and  five ;  872,026 ; 
four  million,  two  thousand,  three  hundred  and  forty- 
seven.  Ans.  5851840. 

34.  Add  eight  hundred  and  eighty  millions,  eight 
hundred  and  eighty-nine  ;  2,002,002  ;  seventy-seven 
million,  four  hundred  and  thirty-six  thousand ;  two  hun- 
dred and  six  million,  five  thousand,  two  hundred  and 
seven  ;  49,003 ;  nine  hundred  and  ninety  million,  nine- 
teen thousand,  nine  hundred  and  nineteen. 

Ans.  2155513020. 


III.  SUBTRACTION. 


Art.  23.  1.  If  you  have  9  apples,  and  give  4  away, 
how  many  are  left?  Ans.  5.  Why?  Because  4  and 
5  arc  9. 

2.  Frank  had  15  cents;  after  spending  7,  how  many 
were  left?     Ans.  8.     Why? 

3.  If  you  take  8  from  13,  how  many  are  left?  Ans.  5. 

The  operation  in  the  preceding  examples  is  termed 
Subtraction.  Hence,  Subtraction  is  the  process  of  Jinding 
the  difference  between  two  numbers. 


SUBTRACTION    OF    SIMPLE    NUMBERS. 


27 


The  larger  number  is  called  the  Minuend ;  the  less,  the 
Subtrahend;  and  the  number  left  after  subtraction,  the 
Difference  or  Remainder. 

Remarks, — 1.  The  word  Minuend  means,  to  he  diminished; 
Subtrahend,  to  be  subtracted. 

2.  In  Addition  (See  Art.  20),  numbers  of  the  same  kind  are 
added;  in  Subtraction  they  are  taken  from  each  other;  therefore, 
Subtraction  is  the  reverse  of  Addition. 

3.  When  the  given  numbers  are  of  the  same  denomination,  the 
operation  is  called  S'.nple  Subtraction. 

Art.  24.  The  Sign  — ,  is  called  minus,  meaning  less. 
Placed  between  two  numbers,  it  denotes  that  the  one  on 
the  ri^ht  is  to  be  taken  from  that  on  the  left. 

Thus,  8  —  5  =  3,  shows  that  5  is  to  be  taken  from  8 ; 
it  is  read,  8  minus  5  equals  3.  Here,  8  is  the  minuend, 
5  the  subtrahend,  and  3  the  remainder. 

SUBTRACTION    TABLE. 


2   2  =  0 

3   3-0 

4   4--  0 

5   5  =  0 

3   2  =  1 

4   3-1 

5   4-1 

6   5  =  1 

4   2-2 

5   3  =  2 

6   4-2 

7   5  =  2 

5   2-3 

6   3-3 

7   4-3 

8   5-3 

6   2  =  4 

7-3-4 

8   4  —  4 

9   5-4 

7   2-5 

8   3-5 

9  —  4  —  5 

10   5  =  5 

8   2-6 

9   3  —  6 

10  —  4-6 

11   5-6 

0   2-7 

10   3  —  7 

11   4  —  7 

12   5-7 

10   2-8 

11   3-8 

12   4  =  8 

13   5-8 

11   2-9 

12   3  —  9 

13   4  -  9 

14   5-9 

6   6-0 

7   7-0 

8   8-0 

9   9-0 

7   6-1 

8   7  =  1 

9   8-1 

10   9-1 

8   6  =  2 

9   7-2 

10   8-2 

11   9  —  2 

9   6-3 

10   7-3 

11   8-3 

12   9-3 

10   6-4 

11   7-4 

12   8-4 

13   9-4 

11   6-5 

12   7-5 

13   8-5 

14   9-5 

12  6-Q 

13   7-6 

14   8   6 

15   9-6 

13   6  —  7 

14  —  7-7 

15  —  8-7 

16   9  —  7 

14   6-8 

15  —  7-8 

16   8   8 

17  —  9  =  8 

15   6-9 

16  —  7  =  9 

17   8   9 

18  —  9-9 

28  RAYS   PRACTICAL  ARITHMETIC. 

Art.  25.  When  numbers  are  small,  the  difference  be- 
tween them  may  be  ascertained  in  the  mind;  when  large, 
the  operation  is  most  easily  performed  by  writing  them. 

EXAMPLES. 

1.  A  man  having  S135,  spent  $112  :  what  sum  had 
he  left? 

Since  only  things  of  the  same  unit  value  can  be  added 
(Art  20),  the  diffe'i-ence  between  things  of  the  same  unit 
value  only  can  be  found ;  hence, 

Place  units  under  ?mtVs,  tens  under  tens^  &c.,  that  the 
fiirures  between  which  the  subtraction  is  to  be  made,  may 
be  in  the  most  convenient  position. 

Solution. — After  arranging  the  numbers,  -f  m  2 

say  2  (units)  from  5  (units)  leave  3  (units),  ^  -2  a 

which  put  in  units'  place ;  then  1  (ten)  from  1^0  minuend. 

3  (tens)  leaves  2  (tens),  which  put  in  tens'  1  1^  subtrahend, 

place  ;  then  1  (hundred)  from  1   (hundred)  2  3  remainder, 
leaves  0,  and  there  being  no  tigures  on  the 

left  of  this,  the  place  is  vacant ;  hence  the  number  of  dollars  left 
is  23  ;  that  is,  135—112  =  23. 

2.  A  farmer  having  2-45  sheep,  sold  123 :  how  many 
had  he  left?  Ans.  122. 

3.  A  man  bought  a  farm  for  $751,  and  sold  it  for 
$875  :  how  much  did  he  gain?  Ans.  $124. 

What  is  the  difference 

4.  Between       73-4  and  531  ?     .     .     .     Ans,       203. 

5.  Between  8752  and  3421?  .  .  .  Ans.  5331. 
C.  Between  529  and  8?  ...  Ans.  521. 
7.  Between  79484  and  25163?      .     .     Ans.  54321. 

Review.— 23.  What  is  Subtraction?  Give  an  example.  What  is  the 
minuend?  The  subtrahend?  Remainder?  Rem.  What  does  minuend 
mean  ?     What  subtrahend  ? 

23.  Why  is  subtraction  the  reverse  of  addition?  24.  What  does  the 
sign  minus  mean  ?  When  placed  between  two  numbers  wha  t  does  it  denote  ? 


SUBTRACTION    OF   SIMPLE   NUMBERS.  29 

Art.  26.  When  the  lower  figure  in  each  order  is  not 
greater  than  the  upper,  the  less  is  subtracted  from  the 
greater,  and  the  difference  written  beneath  ;  but. 

When  the  lower  figure  in  any  order  is  greater  than  the 
upper,  a  difficulty  arises,  which  we  will  now  explain. 

James  had  13  cents ;  after  spending  5,  how  many  had 
he  left? 

5  can  not  be  subtracted  from  3,  but  can  be  from            5 
13;  5  from  13  leaves  8;  hence  he  had  8  left  

Q 

1.  From  73  subtract  45. 

Solution. — Here,  5  units  can  not  be  taken  from 
8  units.  Take  1  (ten)  from  the  7  (tens),  and  add 
this  1  (ten)  or  10  units  to  the  3  units,  -which  will 
make  13  units  in  the  units'  place;  then,  subtract 
the  6  units,  and  there  will  remain  8  units,  to  be 
put  in  units'  place.  Since  1  ten  is  taken  from  the 
7  tens,  there  remain  6  tens  in  the  tens'  place.  Sub- 
tract 4  tens  from  6  tens  and  put  the  remainder  in 
tens'  place.  The  difference  of  the  two  numbers  is 
thus  found  to  be  2  tens  and  8  units,  or  28.  '^  ^ 

Instead  of  actually  taking  1  ten  from  the  7  tens,  and 
adding  it  to  the  3  units,  as  is  done  in  the  margin,  the 
operation  is  performed  in  the  rnind :  thus, 

5  from  13  leaves  8,  and  4  from  6  leaves  2. 

In  such  cases,  the  value  of  the  upper  number  is  not 
changed,  since  the  1  ten  which  is  taken  from  the  order 
of  tens  is  added  to  the  number  in  the  order  of  units. 

EXPLANATIONS. 

Taking  a  unit  of  a  higher  order  and  adding  it  to  the 
units  of  the  next  lower,  so  that  the  figure  beneath  may  be 
subtracted  from  the  sum,  is  called  horrowing  ten. 

After  increasing  the  units  by  10,  instead  of  considering 
the  next  figure  of  the  upper  number  as  diminished  by  1, 

Review. — 25.  When  the  numbers  are  small,  how  is  their  difference 
easily  found  ?  "When  they  are  large  ?  In  writing  numbers  for  subtrac- 
tion, why  place  units  under  units,  tens  under  tens  ? 


T.  V. 

73 

45 

Dif. 

28 

tens. 

units. 

6 

13 

4 

5 

30  RAY'S  PRACTICAL   ARITHMETIC. 

the  result  will  be  the  same,  if  the  next  figure  of  the  lower 
number  be  increased  by  1. 

Thus,  in  the  previous  example,  instead  of  diminishing 
the  7  tens  by  1,  add  1  to  the  4  tens,  which  makes  5  ; 

5  from  7  leaves  2,  the  same  as  4  from  6. 

Hence,  when  a  figure  in  the  lower  number  is  greater 
than  that  above  it,  add  10  to  the  upper  figure,  then  sub- 
tract the  lower  figure  from  the  sum ;  and. 

To  compensate  for  the  10  added  to  the  upper  figure, 
increase  the  next  lower  figure  by  1. 

This  process  depends  on  the  principle,  that  if  any  two 
numbers  he  equally  increased^  their  difference  will  remain 
the  same. 

The  10,  added  to  the  upper  number,  is  equal  to  1  of 
the  next  higher  order  added  to  the  lower  number ;  ten 
units  of  any  order  being  always  equal  to  1  of  the  order 
next  higher. 

2.  Find  the  difi'erence  between  805  and  637. 

Solution,    (1st    Method.) — Writing     the    less  805 

number  under  the  greater,  with  units  of  the  same  6  3  7 

order  under  each  other,  it  is  required  to  subtract  ^  ^  ^ 

the  7  units  from  5  units,  which  is  impossible.  j.  u  o 

The  five  can  not  be  increased  by  borrowing 
from  the  next  figure,  because  it  is  0 ;  therefore,  borrow  1  hun- 
dred from  the  8  hundreds,  which  leaves  7  hundreds  in  hundreds' 
place;  this  1  hundred  makes  10  tens;  then,  borrowing  1  ten  from 
the  10  tens,  and  adding  it  to  the  5  units,  9  tens  will  be  in  the  tens' 
place,  and  15  units  in  the  units'  place. 

Subtracting  7  units  from  15  units,  8  units  are  left,  to  be  written 
in  units'  place  ;  next,  subtracting  3  tens  from  9  tens,  there  are 
left  6    tens,  to  be  written   in    tens'  place;    lastly,   subtracting  6 

Review. — 26.  When  the  lower  figure  is  less  than  the  upper,  how  is  the 
subtraction  performed  ?  What  do  you  understand  by  borrowing  ten  ? 
Illustrate  the  process,  by  subtracting  45  from  73.  After  borrowing  ten, 
what  aiay  be  done  in  order  to  avoid  diminishing  tho  next  upper  figure 
by  1  ?     On  what  principle  does  this  process  depend? 


SUBTRACTION    OF    SIMPLE    NUMBERS.  31 

hundreds  from  7  hundreds,  there  remains  1  hundred,  to  be  written 
iu  hundreds'  place.     The  difference  is  168. 

Or,  (2d  Method.) — If  the  5  units  be  increased  by  10,  say  7 
from  15  leaves  8;  then,  increasing  the  3  by  1,  say  4  from  0  can 
not  be  taken,  but  4  from  10  leaves  6;  then,  increasing  6  by  1, 
say  7  from  8  leaves  1,  and  the  whole  remainder  is  168,  as  before. 

QUESTIONS   TO   BE   SOLVED   BY    BOTH    METHODS. 

3.  From       73  take       48 Ans.     25. 

4.  From     340  take     150 Ans.  190. 

5.  From     508  take     325 Ans.  183. 

6.  From  4603  take  3612 Ans.  991. 

7.  From  8765  take  7766 Ans.  999. 

Remarks. — 1.  The  second  method  is  generally  used ;  it  is  more 
convenient^  and  less  liable  to  error,  especially  when  the  upper 
number  contains  ciphers. 

2.  Begin  at  the  right  to  subtract,  so  that  if  any  lower  figure 
is  greater  than  the  upper,  1  may  be  borrowed  from  a  higher 
order. 

When  each  figure  in  the  lower  number  is  less  than  the  one 
above  it,  the  subtraction  may  commence  at  the  left  hand. 

Art.  27.  If  5  subtracted  from  8  leaves  3,  then  3  added 
to  5  must  produce  8 ;  that  is, 

If  the  difference  of  two  numbers  be  added  to  the  less,  the 
sum  will  be  equal  to  the  greater. 

EULE 

For  Subtraction. — 1.  Wi-ite  the  less  number  under  the 
greater,  placing  units  under  units,  tens  under  tens,  &c. 

2.  Beginning  at  the  right  hand,  subtract  each  figure  from 
the  one  directly  ovei"  it,  and  write  the  remainder  beneath. 

3.  If  the  lower  figure  exceeds  the  dipper,  add  ten  to  the  upper 
figure,  subtract  the  lower  from  it,  and  carry  one  to  the  next 
lower  figure,  or  take  one  from  the  next  upper  figure. 

Proof. — Add  the  remainder  to  the  subtrahend ;  if  the  sum 
is  equal  to  the  minuend,  the  work  is  correct. 

For  proof  by  casting  out  the  9's,  see  "  Rai/'s  Higher  Arithmetic." 


32 


RAY'S   PRACTICAL    ARITHMETIC. 


(8)  (9)  (10)  (11) 

Minuends,         7640       860012       4500120       3860000 
Subtrahends,     1234       430021       2910221         120901 


Remainders, 
Proof    .     . 


6406   429991   1589899   3739099 


7640   860012   4500120   3860000 


12.  Take     1234567  from     4444444. 

13.  Take  15161718  from  91516171. 

14.  Take  34992884  from  63046571. 

15.  153425178  —  53845248  = 

16.  100000000—10001001  = 

17.  Take  17  cents  from  63  cents. 


Ans.  3209877. 
Ans.  76354453. 
Ans.  28053687. 
A71S.  99579930. 
Alls.  89998999. 
Ans.  46  cents. 


18.  A  carriage  cost  $137,  and  a  horse  $65 :  how  much 
more  than  the  horse  did  the  carriage  cost?       Ans.  $72. 

19.  A  tree  75  feet  high  was  broken  ;  the  part  that  fell 
was  37  feet  long :  how  high  was  the  stump  ?  Ans.  38  ft. 

20.  America  was  discovered  by  Columbus  in  1492 : 
how  many  years  had  elapsed  in  1837?  Ans.  345. 

21.  I  deposited  in  the  bank  $1840,  and  drew  out  S475  : 
how  many  dollars  had  I  left?  Ans.  $1365. 

22.  A  man  has  property  worth  810104,  and  owes  debts 
to  the  amount  of  $7426  :  when  his  debts  are  paid,  how 
much  will  be  left?  Ans.  $2678. 

23.  A  man  having  $100000,  gave  away  $11  :  how  many 
had  he  left?  J;<,s.  $99989. 

24.  Subtract  19019  from  20010.  Ans.  991. 

25.  Required  the  excess  of  nine  hundred  and  twelve 
thousand  and  ten,  above  50082.  Ans.  861928. 

26.  Take  4004  from  four  million.  Ans.  3905996. 

27.  Subtract  1009006,  from  two  million,  twenty  thou- 
sand, nine  hundred  and  thirty.  Ans.  1011924. 

Review. — 26.  Rem.  After  borrowing  ten,  which  of  the  two  methods  ia 
generally  used  ?  Why  ?  Why  begin  at  the  right  hand  to  subtract  ? 
27.  Give  the  rule  for  subtraction.     Method  of  proof. 


IV.  MULTIPLICATION. 

Art.  28.  1.  If  1  orange  cost  2  cents,  what  will  3  cost? 

Analysis. — Three  oranges  will  cost  3  times  as  much  as  one. 
That  is,  2  cents  taken  3  times :    2+2+2=6.  Ans. 

2.  If  1  lemon  cost  3  cents,  w^hat  will  4  lemons  cost? 

3+3+3+3  =  12.  Ans, 

3.  In  an  orchard  there  are  4  rows  of  trees,  in  each  row 
21  trees :  how  many  trees  in  the  orchard? 

Solution. — 1st.  By  writing  21  four  times,  igt  j-qw  2  1  trees. 
as  in  the  margin,  and  adding,  the  whole  2d  row  2 1  trees, 
number  of  trees  is  84.  Sd   row,  2  1  t-ees. 

2d.  Instead  of  writing  21  four  times,  write    4th  row    21  trees. 

it  once,  place  the  number  4  under,  it  being  

the  number  of  times  21  is  to  be  taken,  and  84 

say,  4  times  1  (unit)  are  4  (units),  which  put  in  units'  place :  2 1 
then,  4  times  2  (tens)  are  8  (tens),  to  be  put  in  tens' place;  4 
the  result  is  84,  the  same  as  found  by  addition.  ^T 

The  latter  method  is  termed  Multiplication. 

DEFINITIONS. 

Multiplication  is  a  short  method  of  Addition,  w^hen  the 
numbers  to  be  added  are  equal. 

Multiplication  is  also  taking  one  number  as  many  times 
as  there  are  units  in  another. 

The  number  to  be  taken  is  the  multiplicand. 

The  number  denoting  how  many  times  the  multiplicand 
is  taken,  is  the  multiplier.     The  result  is  the  product. 

Thus,  4  times  5  are  20 ;  5  is  the  multiplicand,  4  the 
multiplier,  and  20  the  product. 

The  multiplicand  and  multiplier  are  together  called 
factors,  because  they  make  or  produce  the  product. 

Eeview.' — 28.  What  is  multiplication  ?  What  is  the  multiplicand  ? 
What  does  the  multiplier  denote  ?  What  is  the  product  ?  What  are  the 
multiplicand  and  multiplier  together  called  ?  Why  ?  Kem.  When  is 
multiplication  termed  simple  ? 

3d  Bk.  3  33 


34 


RAY'S   PRACTICAL   ARITHMETIC. 


Remark — When  the  multiplicand  is  of  one  denomination,  the 
operation  is  called  Simple  Multiplication. 

Art.  29.  The  Sign  X ,  read  multiplied  hy,  or  times^ 
denotes  that  the  numbers  between  which  it  is  placed  are 
to  be  multiplied  together. 

Thus,  4X3=12,  shows  that  4  multiplied  by  3,  or  4 
times  3,  are  12,  or  equal  12. 

In  the  table,  the  sign  X ,  may  be  read  times :     thus, 
2  times  2  are  4 ;    2  times  3  are  6 ;    and  so  on. 


MULTIPLICATION    TABLE. 


1X0=0 

2X    0—    0 

3X    0-    0 

4X    0-    0 

1X1=1 

2X    1-    2 

3X    1-   3 

4X    1=   4 

1X2=2 

2X    2        4 

3X    2=    6 

4X    2=    8 

IX    3-    3 

2X    3        6 

3X    3=    9 

4X    3-12 

1X4-4 

2X    4=   8 

3X    4  =  12 

4X    4  —  16 

IX    5-    5 

2X    5-10 

3x    5-15 

4X    5  —  20 

1X6=6 

2X    6-12 

3X    6-18 

4X    6-24 

1X7=7 

2X    7=14 

3x    7-21 

4X    7-28 

1X8=8 

2X    8-16 

3X    8-24 

4-X    8  =  32 

IX    9=    9 

2X    9  =  18 

3X    9-27 

4X    9=36 

1X10-10 

2X  10  =  20 

3X10  =  30 

4X  10-40 

1  X  11  =  11 

2X  11-22 

3X  11-33 

4X  11=44 

1X12  =  12 

2x12-24 

3  X  12-36 

4X  12  =  48 

5x    0-   0 

6X    0-    0 

7X0-0 

8X    0-    0 

5X    1=   5 

6X    1=    6 

7X    1=   7 

8x    1=    8 

5X    2-10 

6x    2-12 

7x    2-14 

8X    2-16 

5X    3  —  15 

6X    3-18 

7X    3-21 

8X    3-24 

5X    4-20 

6X    4  =  24 

7X    4  =  28 

8X    4_32 

5  X    5  =  25 

6X    5-30 

7X    5  =  35 

8X    5  =  40 

5X    0  =  30 

6X    6-36 

7X    6=42 

8X    6-48 

5X    7=35 

6X    7-42 

7X    7=49 

8X    7-56 

5X    8-40 

6X    8  =  48 

7X    8-56 

8X    8-64 

5  X    9-15 

6X    9  =  54 

7X    9_63 

8X    9  =  72 

5x10-50 

6  X  10  — 60 

7X  10-70 

8  X  10-80 

5X11      55 

6X  11—66 

7X  11-77 

8X  11-88 

5  X  12=60 

6  X  12  =  72 

7  X  12  — 84 

8x12  —  96 

MULTIPLICATION    OF   SIMPLE   NUMBERS. 


35 


9X   0—     0 

lOX    0—     0 

11 X    0-     0 

12  X    0—     0 

9X    1—     9 

lOx    1—  10 

llX    1—  11 

12X    1—  12 

9X   2—  18 

lOx    2—  20 

11  X    2—  22 

12  X    2—  24 

9X   3—  27 

lOx    3—  30 

11  X    3—  33 

12  X    3—  36 

9X   4—  36 

10  X   4—  40 

11 X    4—  44 

12  X   4—  48 

9X   5—  45 

lOX    5—  50 

11  X    5==  55 

12X    5—  60 

9X   6—  54 

lOX    6—  60 

11  X    6—  66 

12  X    6—  72 

9X   7—  63 

lOX    7—  70 

11 X   7—  77 

12  X   7=  84 

9X   8—  72 

10  X    8—  80 

11  X   8—  88 

12  X    8—  96 

9X    9—  81 

10  X    9—  90 

11 X    9—  99 

12  X    9—108 

9X10—  90 

10X10^100 

11X10—110 

12x10—120 

9X11—  99 

10x11—110 

llXll— 121 

12X11—132 

9X12—108 

10x12—120 

llXl2:r:=132 

12x12—144 

'K     'P     't^     "1^ 

;!«*** 

'K      '1^      'K      'T" 
JfC     >{C     >fC     ^ 

*  *  *  * 


Art.  30.  Here  are  several  rows  of  stars. 
Counting  upward,  there  are  5  rows  of  4  stars 
each  ;  5  rows  contain  5  times  as  many  as  1  row ; 
5  times  4  stars  are  20  stars. 

Counting  across,  there  are  4  rows  of  5  stars 
each  ;  4  rows  contain  4  times  as  many  as  one 
row ;  4  times  5  stars  are  20  stars,  the  same  as  before. 

Hence,  the  product  of  two  numbers  is  not  altered  hy 
changing  the  order  of  the  factors. 

What  is  the  diiference  between  6X5  and  5X6?  9x8 
and  6X12?     10x12  and  12X10? 

Remark  1. — The  product  is  always  of  the  same  kind  or  denomi- 
nation as  the  multiplicand. 

PRINCIPLES  AND  EXAMPLES. 

1.  What  is  the  product  of  5  multiplied  by  3;  that  is, 
what  is  the  amount  of  5  taken  3  times  ?  Ans.  15. 

3  times  5  are  equal  to  54-5+5=15. 

Here,  the  multiplicand,  5,  is  an  Abstract  number;  that 
is,  it  denotes  no  particular  thing,  and  the  product,  15,  is 
also  abstract. 


Eeviett.— 29.  What  does  the  sign  (  X  )  of  multiplication  denote  ?  Gi\'e 
an  example.  80.  Does  it  alter  the  product  to  make  either  of  the  factors 
the  multiplier  ?    Illustrate  this. 


36  RAY'S  PRACTICAL   ARITHMETIC. 

2.  What  will  3  yards  of  muslin  cost,  at  5  cents  a  yard? 

Analysis. — Three  yards  will  cost  three  times  the  price  of 
one  yard ;  that  is,  5  cents  taken  3  times,  which  is 

5  cts.4-5  cts.-|-5  cts.  =  15  cts.  Ans. 

Here,  the  multiplicand  is  a  Concrete  number  ;  that  is, 
it  denotes  some  particular  thing,  as  Cents;  the  product, 
15  cents,  denoting  the  same  thing,  is  also  concrete. 

Hence,  in  general,  if  the  multiplicand  is  money,  the 
product  will  be  money  of  the  same  name  ;  if  it  is  pounds^ 
the  product  will  be  pounds,  &c. 

Rem.  2. — The  multiplier  shows  the  number  of  times  the  multiplicand  is 
to  be  taken  ;  hence,  it  must  always  be  considered  an  abstract  number. 

In  Example  2,  5  cents  are  not  multiplied  by  3  yards,  but  are  taken 
three  times;  as  three  yards  will  cost  three  times  the  price  of  1  yard. 

To  speak  of  multiplying  8  dollars  by  2  yards,  or  25  cents  by  25  cents, 
is  as  absurd  as  to  propose  to  multiply  3  apples  by  2  ix)tatoes. 

Art.  31.    When  the  Multiplier  does  not  exceed  12. 

EXAMPLES. 

1.  How  many  yards  of  cloth  are  there  in  3  pieces,  each 
containing  123  yards  ? 

Solution. — Having  placed   the   muUi-  operation. 

plier  under  the   multiplicand,  as   in  the        123  multiplicand, 
margin,  say,  3  times  3  units  are  9  units,  3  multiplier. 

which  write  in  units'  place;  3  times  2  tens      ; — 

are  6  tens,  which  write  in  tens'  place ;  3        ^  ^  ^  product, 
times  1  hundred  are  3  hundreds,  which  write  in  hundreds'  place. 

2.  What  will  2  houses  cost  at  $231  each?  Ans.  $462. 

3.  What  will  3  horses  cost  at  $132  each  ?  Ans.  $396. 

4.  What  is  the  product  of    201X4?  Ans.    804. 

5.  What  is  the  product  of  2301x3?  Ans.  6903. 

Review. — 30.  Rem.  1.  What  is  the  denomination  of  the  product  ?  Show 
that  when  the  multiplicand  is  an  abstract  number,  the  product  is  also 
abstract.  Show  that  when  the  multiplicand  is  a  concrete  number,  the 
product  must  be  concrete. 

Rem.  2.  What  does  the  multiplier  show  ?    What  must  it  bo  considered  ? 


MULTIPLICATION    OF    SIMPLE   NUMBERS. 


37 


6.  At  $43  an  acre,  what  will  5  acres  of  land  cost? 

Solution. — Say,  5  times  3  (units)  are  15 
(units) ;  write  the  5  (units)  in  units'  place,  and 
carry  the  1  (ten) ;  then,  5  times  4  (tens)  are  20 
(tens),  and  1  (ten)  carried  make  21  (tens),  which 
write  as  in  the  margin. 

7.  What  is  the  product  of  347  X 12  ? 

The  product  of  each  of  the  figures  by 
12  is  known  from  the  Mul.  Table ;  multi- 
ply then  by  12,  as  by  a  single  figure. 

Proof. — Since  11  times  347  and  1  time 
347  are  equal  to  12  times  347,  therefore, 
take  1  from  the  multiplier,  and  use  the  re- 
mainder instead  of  the  former  multiplier ;  to 
the  product,  add  the  multiplicand ;  the  result 

must  be  the  same  as  the  first  product.  a  ^  ^  . 

^  4164 

RULE 

For  Multiplication. — 1.  Write  the  multiplicand^  with  the 
multiplier  under  it,  and  draw  a  line  beneath. 

2.  Begin  with  units;  multiply  each  figure  of  the  multiplicand 
by  the  multiplier,  and  can^  as  in  addition. 

Proof. — Take  1  from  the  multiplier,  and  use  the  remainder 
for  a  multiplier;  to  the  product  thus  found,  add  the  multipli- 
cand; if  the  sum  equals  the  first  product,  the  work  is  correct 

Rem. — Begin  at  the  right  hand  to  multiply,  so  that  the  excess  of  tens  in 
any  lower  order  may  be  carried  to  the  order  next  higher. 

EXAMPLES  FOR  PRACTICE. 


OPERATION 

843 
5 

8215 

347 
12 

4164 

347 
11 

3817 
347 

Multiplicand, 
Multiplier, 

(8) 

5142 
5 

(9) 
4184 
6 

(10) 

3172 
5 

(11) 
41834 

7 

Product,      .     . 

25710 

25104 

15860 

292838 

Review. — 31.  How  are  the  numbers  written,  Rule  ?  Where  do  you 
begin  to  multiply  ?  Haw  do  you  multiply  ?  How  do  you  prove  the  opera- 
tion ?     Rem.  Why  begin  at  the  right  hand  to  multiply  ? 


38 


RAY'S   PRACTICAL   ARITHMETIC. 


12. 
13. 
14. 
15. 
16. 


EXAMPLES. 

49X3 

57X4 

128X5 

367  X  6 

1427  X  7 


ANS. 

147. 

228. 

640. 
2202. 
9989. 


17. 
18. 
19. 
20. 
21. 


22.  Multiply       46782  by  11. 

23.  Multiply       86458  by  12. 

24.  Multiply     680323  by  11. 

25.  Multiply  1236839  by  12. 


EXAMPLES. 

19645  X  8 

44386  X  9 

708324  X  7 

964578  X  9 

96432  X  10 

Ans. 
Ans. 
Ans. 
Am. 


A\S. 

=  157160. 
=  399474. 
=  4958268. 
=  8681202. 
=   964320. 

514602. 

1037496. 

7483553. 

14842068. 


43 
25 

5 

20 

times 
times 

215  — 

86    — 

43 
43 

1075  =  25  times  43 


Art.  32.    When  the  Multiplier  exceeds  12. 
26.  What  is  the  product  of  43  X  25  ? 

Analysis. — Since  25  is  equal  to  2 
tens  and  5  units,  that  is,  20-|-5, 
multiply  by  5,  and  write  the  product, 
215  (units),  in  units'  place  ;  then 
multiply  by  the  2  tens,  and  set  the 
product,  86  (tens),  in  tens'  place. 

Multiplying    by  5   units  gives  5 
times  43,  and   multiplying   by  2    tens   gives  20  times  43;    add 
them,  because  5  times  43  and  20  times  43:=25X43. 

Hence,  multiply  by  the  units'  figure  of  the  multiplier,  and  write 
the  product  in  units'  place,  because  it  is  units;  multiply  by  the 
tens'  figure,  and  write  the  product  in  tens'  place. 

Tlierefore,  in  multiplying  by  a  figure  of  any  order,  write  the  product 
in  the  same  order  as  the  multiplier. 

Note. — Tho  products  of  the  multiplicand  by  the  separate  figures  of  the 
multiplier,  are  partial  products. 

GENERAL  RULE 

For  Multiplication. — 1.  Write  the  muUipUer  under  the 
muliiplicandj  placing  Jigures  of  the  same  order  under  each  other. 

Review. — 82.  If  tho  partial  multiplier  be  in  units'  place,  where  is  the 
product  written?  Why?  Where  should  the  right  hand  fij^ire  of  each 
product  always  be  written?  Note.  What  are  partial  pmducts?  Give 
the  general  rule  for  multiplication.     What  is  the  method  of  proof? 


MULTIPLICATION   OF  SIMPLE  NUMBERS. 


39 


2.  Multiply  by  each  figure  of  the  rmdiipUer  in  succession, 
beginning  with  imits,  ahoays  setting  the  right  hand  figure  of  each 
product  under  that  figure  of  the  multiplier  which  produces  it. 

3.  Add  the  several  partial  pi'oducts  together ;  their  sum  will 
be  the  product  sought. 

Proof. — Multiply  the  multiplier  by  the  multiplicand  :  the 
product  thus  obtained  must  be  the  same  as  the  tirst  product. 

For  proof  by  casting  out  the  9's,  see  '■'Rays  Higher  Arithmetic.'* 

27.   Multiply  2345  by  123. 

2345  multiplicand. 
123  multiplier. 

7035=       3  times  2345 

4690     =r     20  times  2345 

2345       r=rz  100  times  2345 


PROOF.  123 

2345 

615 

492 

369 

246 

288435  =  123  times  2345 

28.  Multiply  327  by  203. 

When  a  cipher  is  in  the  multiplier,  fill  the  cor- 
responding order  in  the  product  with  0,  or  leave  it 
vacant,  and  multiply  by  the  other  figures. 

Remember  to  place  the  right  hand  figure  of  each 
partial  product  under  the  multiplying  figure. 

By  observing  the  above  principle,  the  multiplica- 
tion may  begin  with  any  figure  of  the  multiplier. 


288435 


327 
203 

981 
654 

66381 


29. 
30. 
31. 
32. 
33. 
34. 
35. 
36. 
37. 


EXAMPLES. 

235  X  13 
346  X  19 
425  X  29 
518  X  34 
279  X  37 
869  X  49 
294  X  57 
429  X  62 
485  X  76 


ANS. 

3055. 
:  6574. 
12325. 
17612. 
10323. 
42581. 
16758. 
26598. 
36860. 


EXAMPLES. 

38.  624  X  85 

39.  976  X  97 

40.  342X364 

41.  376  X  526 

42.  476  X  536 

43.  2187X215 

44.  3489  X  276 

45.  1646X365: 

46.  8432  X  635 


ANS. 

=  53040. 
:  94672. 
:  124488. 
:  197776. 
:  255136. 

470205. 
:  962964. 

600790 
5354320. 


40  RAY'S    PRACTICAL   ARITHMETIC. 

47.  Multiply    6874  by    829.  Ans.      5698546. 

48.  Multiply    2873  by  1823.  Ans.      5237479. 

49.  Multiply    4786  by  3497.  Ans.    16736642. 

50.  Multiply  87603  by  9865.  Ans.  864203595. 

51.  Multiply  83457  by  6835.  Ans.  570428595. 

52.  Multiply  31624  by  7138.  Ans.  225732112. 

53.  "What  will  126  barrels  of  flour  cost,  at  86  a  barrel  ? 

Ans.  $756. 

54.  What  will  823  barrels  of  pork  cost,  at  $12  a  barrel  ? 

Ans.  $9876. 

55.  What  will  675  pounds  of  cheese  cost,  at  13  cents  a 
pound  ?  Ans.  8775  cents. 

56.  What  will  496  bushels  of  potatoes  cost,  at  24  cents 
a  bushel?  Ans.  11904  cents. 

57.  If  a  man  travel  28  miles  a  day,  how  many  miles 
will  he  travel  in  152  days  ?  Ans.  4256  miles. 

58.  There  are  1760  yards  in  one  mile;  how  many  yards 
are  there  in  209  miles?  Ans.  367840  yds. 

59.  There  are  24  hours  in  a  day,  and  365  days  in  a 
year  :  if  a  ship  sail  8  miles  an  hour,  how  far  will  she  sail 
in  a  year?  Ans.  70080  miles. 

60.  Sound  moves  1130  feet  in  a  second:  how  far  will 
it  move  in  109  seconds?  Ans.  123170  feet. 

61.  Multiply  two  thousand  and  twenty-nine  by  one 
thousand  and  seven.  Ans.  2043203. 

62.  Multiply  eighty  thousand  four  hundred  and  one  by 
sixty  thousand  and  seven.  Ans.  4824622807. 

63.  Multiply  one  hundred  and  one  thousand  and  thirty- 
two  by  20001.  Ans.  2020741032. 

CONTRACTIONS    IN    MULTIPLICATION. 

CASE    I. 
Art.  33.    When  the  Multiplier  is  a  Composite  Number. 

Any  number  produced  by  multiplying  together  two  or 
more  numbers,  is  termed  a  composite  number ;   and, 

The  numbers  which,  multiplied  together,  produce  a 
number.,  are  its  component  parts^  or  factors. 


MULTIPLICATION    OF   SIMPLE   NUMBERS.  41 

Thus,  21  is  a  composite  number :  the  factors  are  7  and  3, 
because  7  multiplied  by  3  produces  21.  So,  also,  12  is  a  com- 
posite number;  6X2=12;    or,  3X4=12;    or,  2X2X3=12. 

1.  What  will  15  oranges  cost,  at  8  cents  each  ? 

Since  5X3=15,  it  follows  that  15  is  a  composite  numbei, 
of  which  the  factors  are  5  and  3. 

ANALYsis.-Since  15  are  3  Cost  of    1  orange,  8  cents, 

times  5,  15  oranges  will  cost  c 

3  times  as  much  as  5  oranges.  

Therefore,  instead  of  multi-  Cost  of    5  oranges,.     40  cents. 

plying  8  by  15,  first  find  the  3 

cost  of  5  oranges,  by  multiply-       „  „,,  7~7T7 

Q        .    u     c     ^u        .10       Cost  of  15  oranges,  IJU  cents, 
lug  8  cents  by  5;  then  take  3  ^    ' 

times  that  product  for  the  cost  of  15  oranges. 

Proof.— 8x15=:  120. 

Rule  for  Case  I. — Separate  the  multiplier  into  tico  or  more 
factors.  2.  Multiply  the  multiplicand  hy  one  of  the  factors, 
and  this  product  hy  another  factor^  till  every  factor  is  used; 
the  last  product  will  be  the  one  required. 

Rem, — Do  not  confound  the  factors  of  a  number  with  the  parU 
into  which  it  may  be  separated.  Thus,  the  factors  of  15  are  5  and  3, 
while  the  parts  into  which  15  may  be  separated,  are  any  numbers 
whose  sum  equals  15 ;     as,  14  and  1 ;     or,  2,  9,  and  4. 

EXAMPLES  FOE  PRACTICE. 

2.  "What  will  24  acres  of  land  cost,  at  8124  an  acre  ? 

Ans.  §2976. 

3.  How  far  will  a  ship  sail  in  56  weeks,  at  the  rate  of 
2395  miles  per  week  ?  Ans.  134120  miles. 

4.  How  many  pounds  of  iron  are  there  in  54  loads,  each 
weighing  2873  pounds?  Ans.  155142  pounds. 

5.  How  many  gallons  of  wine  in  63  vats,  each  contain- 
ing 1673  gallons?  Ans.  105399  galls. 

6.  Multiply  2874  by    72.  Ans.  206928. 

7.  Multiply  8074  by  103.  Ans.  871992. 


42  RAYS  PRACTICAL  ARITHMETIC. 

CASE    II. 

Art.  34.  When  the  Multiplier  is  1  icith  ciphers  annexed 
to  it;  as,  10,   100,  1000,  &c. 

Placing  one  cipher  on  the  right  of  a  number,  (Art.  8), 
changes  the  units  into  tens,  the  tens  into  hundreds,  and 
so  on,  and,  therefore,  multiplies  the  number  by  10. 

Annexing  two  ciphers,  changes  units  into  hundreds,  tens 
into  thousands,  &c.,  and  multiplies  the  number  by  100. 

Annex  one  cipher  to  the  right  of  25,  and  it  becomes 
250,  the  product  arising  from  multiplying  it  by  10. 

Annex  two  ciphers  to  25,  and  it  becomes  2500,  the  prod- 
uct of  25  X  100.     Hence  the 

Rule  for  Case  II. — Annex  as  many  ciphers  to  the  Multi- 
plicand as  there  are  ciphers  in  the  multiplier,  and  the  number 
thus  formed  will  be  the  product  required. 

1.  Multiply  245  by        100.  Ans.         24500. 

2.  Multiply  138  by      1000.  "              Ans.       138000. 

3.  Multiply  428  by    10000.  Ans.    4280000. 

4.  Multiply  872  by  100000.  Ans.  87200000. 

CASE    III. 

Art.  35.  When  there  are  ciphers  at  the  right  hand  of 
one  or  both  of  the  factors. 

Find  the  product  of  2300  X  170. 

Analysis. — The  number  2300  may  be  regarded  2300 

as  a  composite  number,  of  which  the  factors  are  23  170 

and  100  i  and  170  as  a  composite  number,  of  which 

the  factors  are  17  and  10.  ^  P  -^ 

By  Art.  33,  the  product  of  2300  by  170  will  be  ^^ 


found  by  multiplying  23  by  17,  and  this  product  by       39100  0 
100,  and  the   resulting   product  by   10;   that  is,  by 
finding  the  product  of  23  multiplied  by  17,  and  then  annexing  to 
the  product  3  ciphers,  as  there  are  3  ciphers  at  the  right  of  both 
factors.     Hence  the 


DIVISION    OF   SIMPLE   NUMBERS. 


43 


Rule  for  Case  111.— Multiphj  without  regarding  the  ciphers 
oil  the  right  of  the  factors  ;  then  annex  to  the  product  as  many 
ciphers  as  are  at  the  right  of  both  factors. 

1.  Multiply  2350  by         60. 

2.  Multiply  80300  by       450. 

3.  Multiply  10240  by     3200. 

4.  Multiply  9600  by     2400. 

5.  Multiply  18001  by  26000. 

6.  Multiply  8602  by     1030. 

7.  Multiply  3007  by     9100. 

8.  Multiply  80600  by     7002. 

9.  Multiply  70302  by  80300. 
10.  Multiply  904000  by  10200. 

For  additional  problems,  see  Ray 


Ans. 

141000. 

Ans. 

36135000. 

Ans. 

32768000. 

Ans. 

23040000. 

Ans. 

468026000. 

Ans. 

8860060. 

Ans. 

27363700. 

Ans, 

564361200. 

Ans. 

5645250600. 

Ans. 

9220800000. 

s  Test 

Examples. 

V.  DIVISION. 

Art,  36.  1.  If  you  divide  6  apples  equally  between  2 
boys,  how  many  will  each  have  ? 

Analysis. — It  will  require  two  apples  to  give  each  hoy  1; 
hence,  each  hoy  will  have  as  many  apples  as  there  are  times  2 
apples  in  6  apples,  that  is  3. 

How  many  times  2  in  6? — Ans.  3.  AYhy?  Because  3 
times  2  are  6. 

2.  If  you  divide  8  peaches  equally  between  2  boys,  how 
many  will  each  have  ?     Ans.  4.     Why  ? 

3.  How  many  times  2  in  10?     Why? 

Review. — 33.  TVTiat  is  a  composite  number  ?  What  arc  its  component 
parts  or  factors  ?  Give  an  example.  How  multiply  by  a  composite  num- 
ber, Rule  ?  Illustrate  this  method.  Rem.  In  what  respect  do  the /actors 
of  a  number  differ  from  its  parts  f     Give  an  example. 

34.  If  one  cipher  is  placed  on  the  right  of  a  number,  how  are  the  orders 
changed?     If  two  ciphers  ?     How  multiply  by  10,  100,  1000,  &c.  ? 


44  RAYS   PRACTICAL   ARITHMETIC. 

Definitions. — The  process  by  which  the  preceding  ex- 
amples are  solved,  is  called  Division ;  hence, 

Division  is  the  process  of  Jinding  how  many  times  one 
number  is  contained  in  another. 

The  number  by  which  to  divide,  is  the  divisor. 
The  number  to  be  divided,  is  the  dividend. 
The  number  denoting   how  many  times  the   divisor  is 
contained  in  the  dividend,  is  the  quotient. 

Art.  37.     How  many  times  3  in  12?     Ans.  4  times. 
Here,  3  is  the  divisor,  12  the  dividend,  and  4  the  quotient. 
Since  3  is  contained  in  12  four  times,  4  times  3  are  12;  that 
is,  the  divisor  and  quotient  multiplied,  produce  the  dividend. 

Hence,  since  3  and  4  are  factors  of  the  product  12,  the  divisor 

and  quotient  correspond  to  the  factors  in  Multiplication ;  the 

dividend,  to  the  product. 

Factors.         Product. 

By  Multiplication,  ...  3  X  4     =12 

Dividend.  Divisor.     Quotient. 

By  Division,     1 2   divided  by  3      =     4 
Or  ...  .     12  divided  by  4     =     3 
Hence,  Division  is  the  process  of  jinding  one  of  the  fac- 
tors of  a  product^  when  the  other  factor  is  known. 

Art.  38.  If  7  cents  be  divided  equally  among  3  boys, 
each  boy  would  receive  2  cents,  and  there  would  be  1  cent 
left,  or  remaining  undivided. 

The  number  left  after  dividing,  is  called  the  remainder. 

Remarks. — 1.  Since  the  remainder  is  a  part  of  the  dividend,  it  must  be 
of  the  same  denomination.  If  the  dividend  be  dollars,  the  remainder  will 
bo  dollars  :  if  pounds,  the  remainder  will  be  pounds. 

2.  The  remainder  is  always  less  than  the  divisor  ;  for,  if  it  were  pqual 
to,  or  greater  than  it,  the  divisor  would  be  contained  at  least  once  more  in 
the  dividend. 

8.  If  the  dividend  denotes  things  of  one  denomination  only,  the  opera- 
tion is  called  Simple  Dwisioji. 

Review. — 36.  What  is  Division?  What  is  the  number  by  which  to 
divide?  What  the  number  to  be  divided?  What  the  number  denoting 
how  many  times  the  divisor  is  contained  in  the  dividend  ? 


8 

cents. 

1st  lemon 

2 

cents. 

Left,  .  .  . 

.  6 

cents. 

2d  lemon 

2 

cents. 

Left,  ... 

.  4 

cents. 

8d  lemon 

2 

cents. 

Left,  .  .  . 

.  2 

cents. 

4th  lemon 

2 

cents. 

Left,  .  .  .  ■ 

,  o" 

cents. 

DIVISION   OF   SIMPLE  NUMBERS.  45 

Art.  39.  A  boy  has  8  cents :  how  many  lemons  can 
he  buy  at  2  cents  each  ? 

Analysis. — He  can  buy  4,  because  4  lemons  at  2  cents  each, 
will  cost  8  cents.  If  he  did  not  know  thai  4  iiines  2  are  8,  the 
operation  would  be  thus  : 

The  boy  would  give  2  cents  for  1 
lemon,  and  then  have  6  cents  left 

After  giving  2  cents  for  the  2d 
lemon,  he  would  have  4  cents  left ; 

Then  giving  2  cents  for  the  third, 
he  would  have  2  cents  left; 

Lastly,  after  giving  two  cents  for 
the  fourth,  he  would  have  nothing 
left:  having  taken  2  cents  4  times 
from  8  cents,  and  each  time  received 
one  lemon. 

The  natural  method  of  performing  this  operation  is  by 
Subtraction ;  but,  ^ 

When  it  is  known  hoio  many  times  2  can  be  subtracted 
from  8,  instead  of  subtracting  2  four  times,  say,  2  in  8 
four  times,  and  4  times  2  are  8 ;  which,  subtracted  from  8 
once,  nothing  is  left. 

The  last  method  is  by  Division,  and  it  differs  from  the 
first  in  this  :  that  the  subtractions,  instead  of  being  per- 
formed separately,  are  all  made  at  once. 

Hence,  Division  may  be  termed  a  short  method  of  mak- 
ing many  i<nhtractions  of  the  same  number. 

The  divisor   is   the   number   subtracted ;    the  dividend 

Review.— 37.  Three  in  12,  4  times  ;  what  is  3  called  ?  12?  4?  To 
what  is  the  product  of  the  divisor  and  quotient  equal  ?  To  what  do  they 
correspond  ?     To  what  does  the  dividend  correspond  ?     What  is  division  ? 

88.  When  there  is  a  number  left  after  dividing,  what  is  it  called? 
Rem.  Of  what  denomination  is  the  remainder  ?  Why  ?  Why  is  the  re- 
mainder always  less  than  the  divisor  ?  When  the  dividend  denotes  things 
of  one  denomination  only,  what  is  the  operation  called  ? 

39.  To  what  natural  method  do  the  operations  in  division  belong  ?  Il- 
lustrate by  an  example.  What  may  division  be  termed  ?  What  is  the 
divisor?     Dividend?     Quotient? 


46 


RAY'S    PRACTICAL    ARITHMETIC. 


the   number  from  which  the  subtractions  are  made  ;  the 
quotient  shows  how  many  subtractions  have  been  made. 


Art.  40.  Division  is  denoted  by  three  Signs: 

1st.     3)12     means  that  12  is  to  be  divided  by  3. 

2d.         -o-      means  that  12  is  to  be  divided  by  3. 

3d.      12-^3  means  that  12  is  to  be  divided  by  3. 

Use  the  Ist  sign  when  the  divisor  does  not  exceed  12 ;  draw 
a  line  under  the  dividend,  and  write  the  quotient  beneath. 

If  the  divisor  exceeds  12,  draw  a  curved  line  on  the  right 
of  the  dividend  :  place  the  quotient  on  the  right  of  this. 

The  sign,  -j-,  in  the  Table,  is  read  divided  by. 


H 


2)8 

15)45(3 

15     .. 

4 

45 

—  o 

5 

21-T-3=7. 


DIVISION    TABLE. 


1^ 

-1=  1 

2- 

f-2-  1 

3fc- 

J-3-  1 

4- 

:  4-  1 

2-i 

-1=  2 

4- 

-2=  2 

6- 

:  3-  2 

8- 

1-4=  2 

3-H 

-1=  3 

6- 

^-2=  3 

9- 

^3-   3 

12- 

1-4=  3 

4-^ 

-1=  4 

8- 

^2=  4 

12- 

f-3-  4 

16- 

^4-  4 

5-, 

-1-  5 

10- 

^2=  5 

15- 

^-3-  5 

20- 

1-4=  5 

6- 

-1-  6 

12- 

1-2=  6 

18- 

f-3-  6 

24- 

:  4-  6 

7- 

:  1—  7 

14- 

^2=  7 

21- 

■-3_  7 

28- 

f-4=  7 

8- 

^1=  8 

16- 

r-2=    8 

24- 

^-3=  8 

32- 

1-4=  8 

9- 

:-l-   9 

18- 

■-2=  9 

27- 

h3-  9 

36- 

-4=  9 

10-; 

1-1-10 

20- 

f-2-10 

30- 

r-3-10 

40- 

-4-10 

5- 

■-5-  1 

6h 

-6=   1 

7- 

-7=   1 

8- 

-8-  1 

10- 

:  5-  2 

12- 

-6=  2 

14- 

7-  2 

16- 

-8=  2 

15- 

:  5-  3 

18- 

-6=  3 

21- 

7-  3 

24- 

-8=  3 

20- 

h^5=  4 

24-- 

-6=  4 

28- 

-7=  4 

32- 

-8=  4 

25- 

1-5=  5 

30h 

-6=  5 

35- 

-7=  5 

40- 

-8=  5 

30- 

:  5=  6 

36- 

-6=  6 

42- 

-7=  6 

48- 

-8=  6 

35- 

-5-  7 

42- 

-6-  7 

49- 

-7-  7 

56- 

-8-  7 

40- 

-5-  8 

48- 

-6-  8 

56- 

-7-  8 

64- 

-8-  8 

45- 

-5-  9 

54- 

-6-   9 

63- 

-7-  9 

72- 

-8-  9 

50- 

-5-10 

60- 

-6=10 

70- 

-7=10    ! 

80- 

-8-10 

DIVISION    OF    SIMPLE    NUMBEKS. 


47 


9--9—   1 

10- 

18  :  9-  2 

20- 

27  :  9-  3 

30- 

36  :  9-  4 

40- 

45  :  9-  5 

50- 

54  :  9—  6 

60- 

63  :  9-  7 

70- 

72  :  9       8 

80- 

81  :  9-  9 

90- 

90  :  9-10 

100-f 

-10=  1 
10=  2 
-10=  3 
10=  4 
10=  5 
10=  6 
10-  7 
-10=  8 
10=  9 
10=10 


11 
22 

33 
44 
55 

66- 

77- 
88- 
99- 
110 


11-  1 

12- 

^12=  1 

11=   2 

24- 

=-12-  2 

11-  3 

36- 

-12-  3 

11—  4 

48- 

-12-  4 

11=   5 

60- 

-12-  5 

11-  6 

72- 

-12—  6 

11-7 

84- 

-12-  7 

11-  8 

96- 

-12-  8 

11=  9 

108- 

-12—  9 

11-10 

120-. 

12—10 

PRINCIPLES  AND  EXAMPLES. 

Art  41.  1.  I  wish  to  put  15  hats  into  boxes,  each  con- 
taining 3  hats :  how  many  boxes  do  I  need  ? 

1st  Solution. — I  need  as  many  boxes     Hats.  Hats, 
as  3  hats  are  coutained  times  in  15  hats;         3)  15  (o,  boxes, 
that  is,  5  boxes.  15 

2.  Having  15  hats,  I  wish  to  separate  them  into  5  equal 
lots  :  how  many  hats  will  there  be  in  each  lot  ? 

2d  Solution. — Putting  one  hat  into  Hats.  Hats, 

each  lot  will  require  5  hats  ;  hence,  *^y  ^  *^  C*^?  ^^*s  m  each. 

there  will  be  as  many  hats  in  each  ^  *^ 
lot,  as  there  are  times  5  hats  in  15. 

The  first  solution  shows,  that  by  Division  a  number  can 
be  separated  into  parts  containing  a  certain  number  of 
units^  and  the  number  of  parts  found. 

The  second  solution  shows,  that  by  Division  a  given 
number  can  be  separated  into  a  certain  number  of  equal 
parts,  and  the  number  of  units  in  each  part  found. 

Review. — 10,  How  many  signs  are  used  to  denote  division  ?  "What  is 
the  first?     Second?     Third?     Illustrate  their  meaning. 

41.  What  does  the  first  solution  show  ?  What  the  second  ?  Eem.  How 
does  it  appear  that  the  divisor  and  dividend  are  both  of  the  same  denomi- 
nation ?  Is  the  quotient  an  abstract  or  a  concrete  number  ?  What  does 
it  show  ?     What  may  it  represent  ? 


48 


RAYS    PRACTICAL   ARITHMETIC. 


MENTAL    EXERCISES. 

Solve  the  examples  on  the  left  like  the  first,  and  thosG  on  the  right,  like 
the  second  of  the  preceding  SoLUTlosa. 


3.  How  many  lemons  at  3 
cents  each,  can  you  buy  for 
12  cents? 

5.  How  many  barrels  of 
flour,  at  4  dollars  a  barrel, 
can  you  purchase  for  $20? 

7.  At  six  dollars  a  yard, 
how  many  yards  of  cloth  can 
you  purchase  for  $30? 

9.  How  many  lead  pencils, 
at  5  cents  apiece,  can  you  buy 
for  35  cents  ? 

11.  A  man  has  63  pounds 
of  butter,  and  wishes  to  put  it 
into  boxes,  each  containing  7 
pounds :  how  many  boxes  will 
be  required  ? 

13.  Into  how  many  parts,  of 
3  each,  can  24  be  separated  ? 


4.  If  you  pay  12  cents  for 
4  lemons,  how  much  will  each 
cost? 

6.  If  you  pay  $20  for  5 
barrels  of  flour,  how  many 
dollars  will  a  barrel  cost  ? 

8.  If  you  pay  $30  for  5 
yards  of  cloth,  how  many  dol- 
lars will  a  yard  cost? 

10.  If  you  pay  35  cents  for 

7  lead  pencils,  how  much  will 
that  be  for  each  ? 

12.  A  man  has  63  pounds 
of  butter,  to  put  into  9  boxes: 
how  many  pounds  must  he 
put  into  each  box? 

14.  If  24  is  separated  into 

8  equal  parts,  how  many  will 
there  be  in  each  part? 


Rem. — 1.  The  divisor  and  dividend  are  both  of  the  same  denomination. 
This  follows  from  that  view  of  division,  which  shows  it  to  be  a  short  method 
t)f  making  several  subtractions  of  the  same  number. 

Since  it  is  only  numbers  of  the  same  denomination  whose  diflforence  can 
be  found,  those  only  of  the  same  denomination  can  be  divided. 

2.  The  quotient  is  an  abstract  number,  and  shows  hoio  many  times  the 
divisor  is  contained  in  the  divided.  But,  it  may  represent  the  number  of 
units  in  some  concrete  number,  as  in  examples  1  and  2,  Art.  41. 


Art.  42.    examples. 
1.  Two  in  6  how  many  times? 


Ans.  3  times. 

2.  How  many  times  is  2  contained  in  GO? 

Analysis. — Two  in  6  units  of  the  first  order,  3 
times:  in  6  units  of  the  second  order,  (10  times  as 
large,)  it  is  contained  10  times  as  many  times. 

3.  How  many  times  is  2  contained  in  600? 


2)60 
3" 


DIVISION    OF    SIMPLE    NUMBERS.  49 

Analysis. — Two  is  contained  3  times  in  6  units  9^6  00 

of  the  first  order ;  but,  in  6  units  of  the  third  order,  

which  are  100  times  as  large,  it  is  contained  100  300 
times  as  often ;  that  is,  300  times. 

4.  Three    feet    make    one    yard :  how    many    yards 
in  60  feet?  Ans.  20. 

5.  Two    pints    make    one    quart :  how    many    quarts 
in  400  pints  ?  Ans.  200. 

6.  How  many  times  3  in  6000?  Ans.  2000. 

7.  How  many  times  4  in  80000?  Ans.  20000. 

Art.  43.     1.  How  many  times  is  2  contained  in  468? 

Here,    the   dividend   is   composed   of  3    numbers;    4   hundreds, 
6  tens,  and  8  units ;  that  is,  of  400,  60,  and  8. 

Divisor.       Dividend.  Quotient. 

Now,      2     in     400     is  contained     200     times, 
2     in       60     ..  ..  30     times. 

2     in         8     ..         ..  4     times. 


Hence,   2     in     468     is  contained  234     times. 

The    same    result    can    be    obtained  without    actually 
separating  the  dividend  into  parts : 

Thus,  2  in  4  (hundreds)  2  (hundred)  times.  Dividend, 

which  write  in  hundreds'  place;    then,  2  in  Divisor   2)468 

G  (tens),  3  (tens)  times,  which  write  in  tens'  .         ~- — 

place;  then,  2  in  8  (units),  4  (uuits)  times,  Quotient  234 
which  write  in  units'  place. 

2.  How  many  times  3  in  693?  An.'i.  231. 

3.  How  many  times  4  in  848?  Ans.  212. 

4.  How  many  times  2  in  4682?  Ans.  2341. 

5.  How  many  times  4  in  8408?  Ans.  2102. 

6.  How  many  times  3  in  36936?  Ajis.  12312. 

7.  How  many  times  2  in  88468  ?  Ajis.  44234. 


Review. — 43.  Explain  how  many  times  2  is  contained  in  468. 
3(1   Bk.  4 


50  RAY'S  PRACTICAL   ARITHMETIC. 

Art.  44.     SHORT    DIVISION. 

1.  How  many  times  is  3  contained  in  129? 

Solution. — Here,  3  is  not  contained   in  1;    but     3)129 

consider  the  1  and  2  as  forming  12  tens,  then  3  is         

in    12   tens,   4   (tens)    times,  which  write   in    tens'  4  3 

place;  3  in  9  (units),  3  times,  which  write  in  units'  place. 

2.  How  many  times  is  3  contained  in  735  ? 

Solution. — Here,  3  is  contained  in  7  (hundreds),      3  ")  7  3  5 

2  (hundred)  times,  and  1  (hundred)  over;  the  1  hun-         

dred  united  with  the  3  tens,  makes  13  tens,  in  which  2  4  5 

3  is  contained  4  (tens)  times  and  1  (ten)  left;   this  1  ten  united 
with  the  5  units,  makes  15  units,  in  which  3  is  contained  5  times. 

OF   REMAINDERS. 

3.  How  many  times  is  3  contained  in  743  ? 

After  dividing,  there  is  2  left,  which  o)7  43 

ought  to  be  divided  by  the  divisor  3:  2  4  7. ..2  Rem. 

But  the  method  of  doing  this  will  not  be  explained 
until  the  pupil  has  studied  Fractions 

The  division  is  merely  indicated  by  placing  the  divisor 
under  the  remainder,  thus,  f . 

The  quotient  is  written  thus,  247f ;  read,  247,  and  two 
divided  hy  three ;  or,  247,  with  a  remainder^  tico. 

Instead  of  performing  all  the  operation  mentally,  the  work 
may  be  written,  as  in  the  following  solution: 

Solution*. — In  this  operation,  say  3  in  7  operation. 

(hundreds),  2  (hundred)  times;  then  multi-  3)743(247f. 

ply  3  by   2   (hundreds),   and    subtract   the  6 

product,  6   (hundreds),  from   7   (hundreds),  "~ — 

which  leaves  a  remainder,  1  (hundred);  to  ^"^* 

this  remainder  unite   the  4  (tens),  making  __!_ 

14  (tens),  which  contains  3,  4  (tens)  times,  23  units, 

with  a  remainder  2  (tens).  21 

To  this  rem.  unite  the  3  units,  and  3  in  23  

(units),  7  times,  with  a  remainder  2.  2  Rem. 


DIVISION    OF    SIMPLE   NUMBEKS.  51 

Definitions. — When  the  division  is  performed  men- 
tally, and  merely  the  result  written,  it  is  termed  Short 
Division;  when  the  entire  work  is  written,  Long  Divi- 
sion. Short  Division  is  used  when  the  divisor  does  not 
exceed  12. 

4.  How  many  times  3  in    462?     .     .     .     Ans.  154. 

5.  How  many  times  5  in  1170?     .     .     .     Aiis.  234. 

6.  How  many  times  4  in     948  ?     .     .     .     Ann.  237. 

Art.  45.     RULE 

For  Short  Division.—!.  Write  the  divisor  at  the  left  of  the 
dividend^  with  a  curved  line  between  them.  Begin  at  the  left  handj 
divide  successively  each  figure  of  the  dividend  hy  the  divisor^ 
and  write  the  result  in  the  same  order  in  the  quotient. 

2.  If  there  is  a  remainder  after  dividing  any  figure.,  prefix  it  to 
the  figure  in  the  next  lower  order,  and  divide  as  before. 

3.  If  the  number  in  any  order  does  not  contain  the  divisor, 
place  a  cipher  in  the  same  order  in  the  quotient,  prefix  the  number 
to  the  figure  in  the  next  lower  order,  and  divide  as  before. 

4.  If  there  is  a  remainder  after  dividing  the  last  figure,  place 
the  divisor  under  it,  and  annex  it  to  the  quotient. 

Explanation. — To  Prefix,  means  to  place  before,  or  at  the  left  hand. 
To  Annex,  is  to  place  after,  or  at  the  right  hand. 

Proof. — Multiply  the  quotient  by  the  divisor,  add  the  re- 
mainder, if  any,  to  the  product:  if  the  work  is  correct,  the 
sum  will  be  equal  to  the  dividend. 

Eeh. — This  method  of  proof  depends  on  the  principle,  Art.  37,  that  a 
dividend  is  a  product,  of  which  the  divisor  and  quotient  are  factors. 

If  the  remainder  be  subtracted  from  the  dividend,  and  the  result  divided 
by  the  quotient,  the  quotient  thus  obtained  will  be  the  divisor. 

Eeview. — 44.  How  many  times  3  in  129  ?  Explain  it.  How  is  a  re- 
mainder to  be  written  ?  "When  is  the  operation  termed  short  division  ? 
When  long  division  ?  When  is  short  division  used  ?  When  lonoj  ?  What 
is  the  diflFerence  between  long  and  short  division  ? 

45.  In  dividing,  how  are  the  numbers  written,  Eule  ?  Where  do  you 
begin  to  divide  ?    Where  is  each  quotient  figure  placed  ? 


52  RAYS    PRACTICAL   ARITHMETIC. 

7.  Divide  653  cents  by  3. 


Dividend. 
Divisor  .  .  3)653 

Quotient.  .  .  217  Rem.  2. 

(8)        (9) 
6)454212   7)874293 


217   PROOF. 

3 


651=  cents  divided. 
2  =  remainder. 


6  5  3  =  the  dividend. 

(10)       (11) 
9)2645402   8)3756031 


Ans. 


75702 
6 


Proof.  454212 


124899 

7 

874293 


2939335 
9 

2645402 


4695031 
8^ 

3756031 


OF   PARTS   OF   NUMBERS. 

Note. — When  any  number  is  divided  into  two  equal  parts, 
one  of  the  parts  is  called  one-half  of  that  number. 

If  divided  into  three  equal  parts,  one  of  the  parts  is  called 
one-third.  If  into  four  equal  parts,  one-fourth.  If  into  five 
equal  parts,  one-fifth;  and  so  on.     Hence, 

To  find  one-half  of  a  number,  divide  by  2 ;  to  find  one-thirdy 
divide  by  3 ;  one-fourth^  divide  by  4 ;     one-fifth,  by  5,  &c. 


12.  Divide 

13.  Divide 

14.  Divide 
Divide 
Divide 
Divide 


15 
16 

17 

18 


8652  by  2. 
406235  by  3. 
675043  by  4. 
984275  by  5. 
258703  by  6. 
523408  by  6. 


Divide  8643275  by  7. 


Ans. 

4326. 

Ans. 

1354112. 

Ans. 

168760f. 

Ans. 

196855. 

Ans. 

43117J. 

Ans. 

87234.^. 

Ans.  12347534. 


Review. — 45.  II  iw  do  you  proceed  when  there  is  a  remainder,  after 
dividing  any  order?  "When  tho  number  in  any  order  does  not  contain  tho 
divisor?  When  there  is  a  remainder  after  dividing  tho  last  figure  ?  What 
is  the  method  of  proof  ?  Rem.  Upon  what  principle  docs  tho  proof  depend  ? 
What  other  method  of  proof  ? 

45.  Note.  When  you  divide  a  number  into  two  equal  parts,  what  is  ono 
part  called?  If  into  three  equal  parts,  what  is  ono  part  called?  If  into 
four  equal  parts  ?  If  into  five  equal  parts  ?  How  do  you  find  one-half  of 
a  number?  One-third?  One-fourth?  One-fifth?  When  a  number  ia 
divided  by  Y,  (example  18,)  what  part  of  tho  number  is  found  ? 


DIVISION   OF    SIMPLE   NUMBERS. 


53 


19. 

Divide     6032520  by 

8. 

Ans. 

754005. 

20. 

Divide     9032706  by 

9. 

Ans. 

1003634. 

21. 

Divide     1830024  by 

10. 

Ans. 

183002^4^. 

22. 

Divide       603251  by 

11. 

Ans. 

54841. 

23. 

Divide  41674008  by 

12. 

Ans. 

3472834. 

24.  If  oranges  cost  3  cents   each,  how   many  can  be 
bought  for  894  cents  ?  Ans.  298  oranges. 

25.  If  4  bushels  of  apples  cost  140  cents,  how  much  is 
that  a  bushel  ?  Ans.  35  cts. 

26.  If  flour  cost  ^4  a  barrel,  how  many  barrels  can  be 
bought  for  $812  ?  Ans.  203  barl. 

27.  A  carpenter  receives  ^23  for  9  months'  work  :  how 
much  is  that  per  month  ?  Ans.  $47. 

28.  There  are  12  months  in  1  year :  how  many  years 
are  there  in  540  months?  Ans.  45  yrs. 

29.  There  are  4  quarts  in  1  gallon :  how  many  gallons 
are  there  in  321276  quarts?  Ans.  80319  gals. 

30.  At  $8  a  barrel,  how  many  barrels  of  flour  can  be 
bought  for  $1736  ?  Ans.  217  barl. 

31.  There  are  7  days  in  one  week  :  how  many  weeks  are 
there  in  734566  days  ?  Ans.  104938  wks. 

32.  A  number  has  been  multiplied  by  11,  and  the  prod- 
uct is  495:  what  is  the  number?  Ans.  45. 

33.  The  product  of  two  numbers  is  3582 :  one  of  the 
numbers  is  9  :  what  is  the  other  ?  Ans.  398. 


34. 

Find 

one-half 

of       56. 

.     .     Ans. 

28. 

35. 

Find 

one-half 

of  3725. 

,     .     Ans. 

18621. 

36. 

Find 

one-third 

of    147.       . 

,     .     Ans. 

49. 

37. 

Find 

one -fourth 

of    500. 

,     .     Alls. 

125. 

38. 

Find 

one-fifth 

of  1945. 

,     .     Ans. 

389. 

39. 

Find 

one-sixth 

of  4476. 

,     .     A71S. 

746. 

40. 

Find 

one-seventh 

of  2513. 

,     .     Ans. 

359. 

41. 

Find 

one-eighth 

of  5992. 

Ans. 

749. 

42. 

Find 

one-ninth 

of  8793.       . 

.     Ans. 

977. 

43. 

Find 

one-tenth 

of  1090.       . 

.     Ans. 

109. 

54  RAY'S   PRACTICAL    ARITHMETIC. 

44.  Find  one-eleventh  of  4125.     .     .     .     Ans.  375. 

45.  Find    one-twelfth    of  5556.     .     .     .     Ajis.  463. 

46.  I  divided  144  apples  equally  among  4  boys ;  the 
eldest  boy  gave  one-third  of  his  share  to  his  sister :  what 
number  did  the  sister  receive  ?  Ans.  12. 

47.  James  found  195  cents,  and  gave  to  Daniel  one- 
fifth  of  them :  Daniel  gave  one-third  of  his  share  to  his 
sister:  how  many  cents  did  she  receive?  Ans.  13. 

Art.  46.    LONG    DIVISION. 

1.  Divide  3465  dollars  equally  among  15  men. 

Solution, — When  the  divisor  exceeds  12,  it  becomes  necessary 
to  write  the  process,  as  in  "operation,"  Example  3,  page  50. 

Fifteen  is  not  contained  in  3  (thousands),  therefore,  there  will 
be  no  thousands  in  the  quotient.  Take  34  (hundreds)  as  a  partial 
dividend;  15  is  contained  in  it  2  (hundred)  times;  that  is,  15  men 
have   each  $200,  which  requires  in  all  «••„•• 

15  X  2  (hundred)  =  30  hundreds.  1 1  I  "3    1  I  I 

Subtract  30  (hundreds)  from  34  (huii-  15)3465(231 
dreds)  and  4  (hundreds)  remain,  to  which  3  0  huud. 

bring  down  the  6  (tens),  and  you  have  4  n 

46  (tens)  for  a  second  partial  dividend.  ,  - 

46  contains  15,  3  (tens)  times,  giving  

each  man  3  ten  (30)  dollars  more,  requiring  ^^  units, 

for  all,  16X3  (tens)  =45  tens  of  dollars.  1^ 

Subtract  45  and  bring  down  the  5  (units),  you  have  15  (units) 
for  a  3d  partial  dividend,  in  which  the  divisor,  15,  is  contained 
once,  giving  to  each  man  1  (unit)  dollar. 

Uencc^  each  man  receives  2  hundred  dollars,  3  ten  doK 
lars,  and  1  dollar ;  that  is,  231  dollars. 

By  this  process,  the  dividend  is  sep-  Divisor.    Parts.  Quotients, 

arated  into  parts,  each  containing  the  15       3000  200 

divisor  a  certain  number  of  times.  4  5  0  3  0 

The  first  part,  30  (hundreds),  con-  15  1 

tains  the  divisor  2  (hundred)  times;  «  «  />  k  9  ':i.-\ 
the  second  part,  45  (tens),  contains  it  3 

(ten)  times;  the  third  part,  15  (units),  contains  it  1  (unit)  time. 


DIVISION  OF  SIMPLE    NUMBERS.  55 

The  several  parts  together,  equal  the  given  dividend:  the 
several  partial  quotients  make  up  the  entire  quotient. 

2.  In  147095  days,  how  many  years,  each  365  days  ? 

Solution.  —  Taking  147  365)14  7  095(403  years, 
(thousand)  for  the  first  par-  14  6  0 

tial  dividend,  we  find  it  will  ^  (\  Q  " 

not  contain  the  divisor;  hence  1  0  Q  ^i 

use  four  figures.  

Again,  after  multiplying  and  subtracting,  as  in  the  preceding  ex- 
ample,andbringingdown  the9  (tens),  thepartialdividend,  109  (tens), 
will  not  contain  the  divisor;  hence,  write  a  0  (no  tens)  in  the  quotient, 
and  bring  down  the  5  (units) :  the  last  partial  dividend  is  1095 
(units),  which  contains  the  divisor  3  (units)  times. 

3.  Divide  4056  by  13 .     Ans.  312. 

RULE 

For  Long  Division. — I.  Place  the  divisor  on  the  left  of  the 
dividend,  draw  a  curved  line  between  them,  and  another  on  the 
right  of  the  dividend, 

2.  Find  how  many  times  the  divisor  is  contained  in  the  fewest 
left  hand  figures  of  the  dividend  that  will  contain  the  divisor^ 
and  place  this  number  in  the  quotient  at  the  right. 

3.  Multiply  the  divisor  by  this  quotientfgure ;  place  the  product 
under  that  part  of  the  dividend  from  which  it  was  obtained. 

4.  Subtract  this  product  from  the  figures  above  it ;  to  the  re- 
mainder bring  down  the  next  figure  of  the  dividend^  and  divide  as 
before,  until  all  the  figures  of  the  dividend  are  brought  down. 

5.  If  at  any  time,  after  bringing  down  a  figure,  the  number 
thus  formed  is  too  small  to  contain  the  divisor,  place  a  cipher 
in  the  quotient,  and  bring  down  another  figure,  after  which 
divide  as  before. Proof,  Same  as  in  Short  Division. 

Eeview. — 46.  Repeat  the  Eule  for  Long  Division.  "Where  is  the  divi- 
sor placed?     Where  the  quotient  ?     How  commence  dividing? 

46.  After  obtaining  the  first  quotient  figure,  how  proceed  ?  After  bring- 
ing down  a  figure,  if  the  partial  dividend  thus  formed  is  too  small  to  contain 
the  divisor,  what  is  to  be  done  ?     What  the  method  of  proof? 


56 


RAY'S  PRACTICAL  ARITHMETIC. 


Notes. — 1.  The  product  must  never  be  greater  than  the  partial 
dividend  from  which  it  is  to  be  subtracted.  When  so,  the  quotient 
figure  is  too  large,  and  must  be  diminished. 

2.  After  subtracting,  the  remainder  must  always  be  less  than 
the  divisor.  When  the  remainder  is  not  less  than  the  divisor,  the 
last  quotient  figure  is  too  small,  and  must  be  increased. 

3.  Example  1  shows  that  the  order  of  each  quotient  figure  is  the 
same  as  the  lowest  order  in  the  partial  dividend  from  which  it  was 
obtained.  It  will  be  a  useful  exercise  to  name  the  order  of  each 
quotient  figure  immediately  after  obtaining  it. 

For  casting  out  the  9'8,  see  '^^  Ray's  Higher  Arithmetic.^^ 


4.  Divide  78994  by  319. 


Prooi 

247  Quotient. 

319)78994 

y-»  i-k  /-\ 

K247 

319  Divisor. 

bd« 

2223 

1519 

247 

1276 

— 

— 

741 

78793 

2434 

2233 

Rem. 

Add 

201  Re 
78994  = 

mainder. 

201 

the  Dividen 

5. 

Divide 

11577 

by 

14.     .     . 

.     Ans. 

826 if. 

6. 

Divide 

48690 

by 

15.     . 

.     Ans. 

3246. 

7. 

Divide  1110960 

by 

23.     . 

.     Atis. 

48302A|. 

8. 

Divide 

122878 

by 

67.     . 

,        A71S. 

1834. 

9. 

Divide 

12412 

by 

53.     . 

,     A  ns. 

23410. 

10. 

Divide 

146304 

by 

72.     .     . 

Ans. 

2032. 

11. 

Divide 

47100 

by 

54.     . 

.     Ans. 

8724f. 

12. 

Divide 

71104 

by 

88.     . 

.     A71S. 

808. 

13. 

Divide 

43956 

by 

66.     . 

.     Ans. 

666. 

14. 

Divide 

121900 

by 

99.     . 

.     Ans. 

123131. 

15. 

Divide 

25312 

by 

112.     . 

,     Ans. 

226. 

16. 

Divide 

381600 

by 

123.     . 

A71S. 

3102^V 

17. 

Divide 

105672 

by 

204.     .     . 

Ans. 

518. 

18. 

Divide 

600000 

by 

1234.    .     . 

A71S. 

486,^4- 

DIVISION    OF    SIMPLE    NDxMBERS.  67 


19. 

Divide 

47263488  by 

4674. 

Ans.   10112. 

20. 

Divide 

26497935  by 

2034. 

Ans.   13027J-g^|. 

21. 

Divide 

48905952  by 

9876. 

Ans.   4952. 

22. 

Divide 

4049160  by 

12345. 

Ans.     328. 

23.  Divide  552160000  by       973.      Aiis.  567482yV3. 

24.  At  815   an  acre,   how  many   acres   of  land  can  be 
bought  for  §3465  ?  Ans.  231  acres. 

25.  If  a  man  travel  26  miles  a  day,  in  how  many  days 
will  he  travel  364  miles  ?  Ans.  14  days. 

26.  If  §1083  be   divided   equally  among  19  men,  how 
many  dollars  will  each  have  ?  Ans.  §57. 

27.  A  man  raised  9523  bushels  of  corn  on  107  acres : 
how  much  was  that  on  one  acre  ?  Ans.  89  bush. 

28.  In   1   hogshead  there   are   63  gallons :  how  many 
hogsheads  in  14868  gallons  ?  Ans.  236  hds. 

29.  A  President  receives  §25000   a  year   (365   days)  : 
how  much  is  that  a  day  ?    Ans.  §68  a  day,  and  §180  over. 

30.  The  yearly  income  from  a  railroad  is  §37960  :  how 
much  is  that  per  day  ?     (365  days=l  year.)       Ans.  §104. 

31.  The  product  of  two  numbers  is  6571435  ;  one  of 
the  factors  is  1235  :  what  is  the  other?  Ans.  5321. 

32.  Divide  one    million   two   hundred  and   forty-seven 
thousand  and  four  hundred  by  405.  Ans.  3080. 

33.  Divide  10  million  four  hundred  and  one  thousand 
by  one  thousand  and  six.  Ans.  10338yyQ^g. 

CONTRACTIONS    IN    DIVISION. 

CASE    I. 
Art.  47.    When  the  Divisor  is  a  Composite  Number, 

1.  A  man  paid  §255  for  15  acres  of  land  :  how  much 
was  that  per  acre  ? 

Analysis.  —  By    taking  t^  ,, 

•^  =>  Dollars. 

one-third  of  §255,  you  ob-      q\of;k.        xi.         ^         £>  tr. 

,  '  -^     , .  ,       o  )  25  D  ^  the  value  of  15  acres. 


tain  the  value  of  one-third, 
(5  acres, )  of  the  land ;  divi- 
ding this  quotient  (85)  by 
6,  gives  the  value  of  1  acre. 


(5  acres,)  of  the  land;  divi-  5)85  =  the  value  of    5  acres. 

ding  this  quotient  (85)  by  ^  _,  ,  ^ 

c     f      .u        ,        ^■,  1 7  =  the  value  of    1  acre. 


58  RAY'S   PRACTICAL   ARITHMETIC. 

The  preceding  analysis  shows  that  instead  of  dividing  by  the  composite 
number  15,  whose  factors  are  3  and  5,  we  may  first  divide  by  one  factor, 
then  divide  the  quotient  thus  obtained  by  the  other  factor. 

2.  Find  the  quotient  of  37,  divided  by  1-4. 

Solution. — Dividing  by  2,  the  quo-  2^)3  7 

tient  is  18  twos  and  1  unit  remaining.      

Dividing  by  7,  the  quotient  is  2,  with  7)18  and  1  over. 

a  remainder  of  4  twos ;  the  whole  re-  o        •,   *   ,        ,  -. 

...        •    ^         '    ,       1        Q  J  and  4  twos  left, 

mainder  then,  is  4  twos  plus  1,  or  U. 

Rule  for  Case  I. — Divide  the  dividend  by  one  of  the  factors 
of  the  divisor;  then  divide  the  quotient  thus  obtained  by  the 
other  factor. 

2.  Multiply  the  last  remainder  by  the  first  divisor;  to  the 
product  add  the  first  remainder ;  the  amount  will  be  the  true 
remainder. 

Note. — When  the  divisor  can  be  resolved  into  more  than  two  factors, 
you  may  divide  by  them  successively.  The  true  remainder  will  be  found 
by  multiplying  each  remainder  by  all  the  preceding  divisors,  except  that 
which  produced  it.     To  their  sum  add  the  remainder  from  first  divisor. 

3.  Divide     2583  by  63.    (63  =  7X9)  Am.  41. 

4.  Divide     6976  by  32.    (32  =  4x8)  Ans.  218. 

5.  Divide     2744  by  28.    (28  =  7x4)  Ans.  98. 

6.  Divide     6145  by  42.    (42  =  6x7)  Ans.  146.}-i 

7.  Divide  19008  by  132 Ans.  144. 

8.  Divide     7840  by     64 Ans.  122|J. 

9.  Divide  14771  by     72 Ans.  205^^. 

10.  Divide  10206  by     81 Ans.  126. 

CASE    II. 

Art.  48.  To  divide  by  1  with  ciphers  annexed;  as  10, 
100,  1000,  i&c. 

Kevikw. — 40.  Note.  When  any  product  is  greater  than  the  partial 
dividend  from  which  it  is  to  be  subtracted,  what  must  be  done  ? 

47.  How  may  division  be  performed,  when  the  divisor  is  a  composite 
number  ?  How  is  the  true  remainder  found  ?  Note.  When  the  divisor 
can  bo  resolved  into  more  than  two  factors,  how  may  the  division  be 
performed  ?     How  is  the  true  remainder  obtained  ? 


DIVISION   OF   SIMPLE   NUMBERS. 


59 


To  multiply  G  by  10,  annex  one  cipher,  thus,  60. 
On  the  principle  that  division  is  the  reverse  of  multipli- 
cation, to  divide  60  by  10,  cut  off  a  cipher. 

Had  the  dividend  been  65,  the  5  might  have  been  separated  in 
like  manner  as  the  cipher ;  6  being  the  quotient,  5  the  remainder. 
The  same  will  apply  when  the  divisor  is  100,  1000,  &c. 

Rule  for  Case  II. —  Out  off  as  many  figures  from  the  right 
of  the  dividend  as  there  are  ciphers  in  the  divisor ;  the  figures 
cut  off  will  be  the  remainder,  the  other  figures,  the  quotient. 

1.  Divide  34872  by  100. 

OPERATION.    1 1 00)348 1 72 


Divide 
Divide 
Divide 
Divide 
Divide 

7.  Divide 

8.  Divide 


2 
3 
4 
5 
6 


2682 
4700 
37201 
46250 
62034 
18003 
375000 


by  10. 

by  100. 

by  100. 

by  100. 

by  100. 

by  1000. 

by  1000. 


348  Quo.  72  Rem. 


Ans.  26Sj%, 


Ans. 
Ans. 
Ans. 
Alls. 


Ans.   47. 

4-62  5  0 
620  3  4 

Ans.   375. 


CASE  III. 

Art.  49.    To  divide,  when  there  are  ciphers  on  the  right 
of  the  divisor. 

1.  Divide  4072  by  800. 

Solution. — Regard  800  as 
a  composite  number,  the  fac- 
tors 100  and  8,  and  divide  as 
in  the  margin. 

In  dividing  by  800,  sepa- 
rate the  two  right  hand  fig- 
ures for  the  remainder,  then ' 
divide  by  8. 


OPERATION. 

1|00)40|72 
8)40 

5  Quo    72  Rem. 

8|00)40|72 

5  Quo.  72  Rem. 


Review. — 48.  How  do  you  divide  by  10,  100,  1000,  &c.  ?  On  what 
principle  does  the  rule  for  case  II  depend  ?  49.  How  do  you  divide  when 
there  are  ciphers  on  the  right  of  divisor,  Rule  for  case  III  ? 


60 


RAY'S    PRACTICAL    ARITHMETIC. 


59 

48 

"IT 


2.  Divide  77939  by  2400. 

Solution.— Since  2400  equals  24100)779139  (3  2^1^§. 
24X100,  cut  oflF   the   two  right  72 

hand  figures,  the  same  as  divid- 
ing by  100;  then  divide  by  24. 

Dividing  by  100,  the  remainder  is  39;  di- 
viding by  24,  the  remainder  is  11.  To  find 
the  true  remainder,  multiply  11  by  100,  and 

add  39  to  the  product,  (Art.  47,  Rule);   this  is  the  same  as  annex- 
ing the  figures  cut  off",  to  the  last  remainder.     Hence,  the 

Rule  for  Case  III. — 1.  Cut  off  the  ciphers  at  the  right  of  the 
divisor^  and  as  many  figures  from  the  right  of  the  dividend. 

2.  Divide  the  remaining  figures  in  the  dividend  by  the  re- 
maining figures  in  the  divisor. 

3.  Annex  the  figures  cut  off  to  the  remainder^  which  gives  the 
true  remainder. 

3. 
4. 
5. 
6. 

7. 

8. 

9. 

10. 

Exercises  in  more  difficult  contractions  are  in  "  Ray*s  Higher  Arithmetic." 

PROOF  OF  MULTIPLICATION  BY  DlVISIOr 

Art.  50.  Division  (Art.  37),  is  a  process  for  finding 
one  of  the  factors  of  a  product  when  the  other  factor  is 
known  :   therefore, 

If  the  product  of  two  number.<}  be  divided  by  the  multiplier, 
the  quotient  will  be  the  multiplicand :  Or^  if  divided  by  the 
multiplicand,  the  quotient  will  be  the  multiplier. 


Divide        73005  by  4000.     .     .     . 

Ans.    ISigei 

Divide        36001  by  9000.     .     . 

•       -^**^-     4  915  0  0  • 

Divide    1078000  by  11000. 

.      .      .      AnS.dS. 

Divide        40167  by  180.       .     .     . 

Ans.    223i\\. 

Divide      907237  by  2100.     .     .     . 

J.?is.  43277^  Jq. 

Divide      364006  by  6400.     .     .     . 

Ans.    56||og. 

Divide  76546037  by  250000.      . 

Ans.  S06.,%o^%\. 

Divide  43563754  by  63400.    .     . 

Ans.    687gW^^5. 

Review.  —  50.  What  is  division?    If  the  product  of  t>Yo  factors  be 
divided  by  either  of  them,  what  will  bo  the  quotient? 


REVIEW   OF   PRINCIPLES.  61 

1.  What  number  multiplied  by  7895,  will  give  434225 
for  a  product?  Ajis.  55. 

2.  If  327  be  multiplied  by  itself,  the  product  will  be 
106929.     Give  the  proof. 

3.  The  product  is  10741125  ;  the  multiplier  375  :  what 
is  the  multiplicand?  Ans.  28643. 

4.  The    product    is    63550656,    and    the   multiplicand 
60352  :  what  is  the  multiplier?  A}is.  1053. 

E^°  For  additional  problems,  see  Ray"s  Test  Examples. 


REVIEW   OF    PRINCIPLES. 

Art.  51.  Notation  and  Numeration  show  how  to 
express  numbers  by  words,  by  /igiires,  or  by  letters. 

For  other  scales  of  notation  than  the  decimal  or  tens'  scale,  an  interest- 
ing subject  for  advanced  students,  see  '*  Hay's  Higher  Arithmetic." 

Art.  52      BY    ADDITION, 

The  aggregate  or  sum  of  two  or  more  numbers  is  found, 
(Art.  18).  Thus,  when  the  separate  cost  of  several  things 
is  given,  the  entire  cost  is  found  by  addition. 

Example. — A  bag  of  coffee  cost  $23,  a  chest  of  tea  S38, 
a  box  of  sugar  $11 :  what  did  all  cost?  Ans.  $72. 

Art.    53.     BY    SUBTRACTION, 

The  difference  between  two  numbers  is  found.  Thus, 
if  the  sum  of  two  numbers  be  diminished  by  either  of  theni^ 
the  remainder  will  be  the  other. 

Hence,  by  Addition,  if  the  difference  of  two  numbers  be 
added  to  the  less,  the  SUM  will  be  the  greater. 

Eeview. — 51.  What  do  Notation  and  Numeration  show  ?  52.  What  is 
found  by  addition  ?     Give  an  example. 

53.  What  is  found  by  subtraction  ?  Having  the  sum  of  two  numbers, 
and  one  of  them,  how  is  the  other  found  ?  When  the  smaller  of  two  num- 
bers and  the  difference  are  given,  how  is  the  greater  found  ?  When  the 
difference  and  greater  are  given,  how  is  the  less  found  ? 


62  RAY'S   PRACTICAL   ARITHMETIC. 

Example  1.  The  sum  of  two  numbers  is  85  ;  the  less  num- 
ber is  37 :  what  is  the  greater  ? 

2.  The  sum  of  two  numbers  is  85 ;  the  greater  number 
is  48  :  what  is  the  less  ? 

3.  The  difference  of  two  numbers  is  48;  the  less  number 
is  37:  what  is  the  greater? 

4.  The  difference  of  two  numbers  is  48 ;  the  greater  number 
is  85 :  what  is  the  less  ? 

Art.  54.     BY    MULTIPLICATI02T, 

Is  foand  the  amount  of  a  number  taken  as  many  times  as 
there  are  units  in  another.  Art.  28. 

Hence,  having  the  cost  of  a  single  thing,  to  find  the  cost  of 
any  number  of  things,  multiply  the  cost  of  one  by  the  number 
of  things. 

1.  If  1  yard  of  tape  cost  3  cents,  what  will  5  yards  cost? 

Analysis. — Five  yards  are  5  times  1  yard  ;  therefore  5  yards 
will  cost  5  TIMES  as  much  as  1 :  the  entire  cost  is  found  by  mul- 
tiplying the  price  of\  yard  by  the  number  of  yards. 

The  divisor  and  quotient  being  given,  the  dividend  is  found 
by  multiplying  together  the  divisor  and  quotient.  Art.  37. 

2.  A  divisor  is  15 ;  the  quotient  is  12 :  what  is  the 
dividend  ? 

3.  An  estate  was  divided  among  7  children ;  each  child 
received  $525  :  what  sum  was  divided  ?  Ans.  $3675. 

Art.  55.    BY  Division, 

Is  found  how  many  times  one  number  is  contained  in 
another.  Art.  36.     This  enables  us, 

1.  To  divide  any  number  into  parts,  each  part  containing  a 
certain  number  of  units. 

2.  To  divide  a  number  into  any  given  mimber  of  equal  parts. 

Thus,  if  the  cost  of  a  number  of  things  and  the  price  of  one  are 
given,  the  number  of  things  is  found  by  division. 

1.  James  spent  35  cents  for  oranges,  and  paid  5  cents 
each  :  how  many  did  he  buy? 

Solution. — He  got  one  orange  for  each  time  5  cents  are  contained 
in  85  cents;     6  in  35,  7  times;  therefore,  he  bought  7  oranges. 


REVIEW   OF   PRINCIPLES.  63 

Knowing  the  cost  of  a  given  number  of  things,  we  obtain  the 
price  of  one^  by  dividing  the  whole  cost  into  as  many  equal  part3 
as  there  are  things. 

2.  If  4  oranges  cost  20  cents,  what  does  one  cost? 

Solution. — If  20  cents  be  divided  into  4  equal  parts,  each  part 
will  be  the  cost  of  1  orange.  Placing  1  cent  to  each  part,  will  re- 
quire four  cents.  Hence,  there  will  be  as  many  cents  in  each  part, 
as  4  cents  are  contained  times  in  20  cents;  4  in  20,  5  times;  hence, 
in  each  part  there  will  be  6  cents,  the  cost  of  one  orange. 

If  the  product  of  two  factors  be  divided  by  either  of 
them,  the  quotient  will  be  the  other.  Art.  37. 

Hence,  if  the  dividend  and  quotient  he  given,  Jind  the 
divisor  by  dividing  the  dividend  by  the  quotient. 

Thei^efore^  having  the  product  of  three  numbers,  and  tico  of 
them  given,  the  third  can  be  found  by  dividing  the  product  ot  the 
three  numbers  by  the  product  of  the  two  given  numbers. 

3.  A  dividend  is  2875 ;  the  quotient,  125 :  find  the 
divisor.  Ans.  23. 

4.  The  product  of  three  numbers  is  3900 :  one  number 
is  12,  another  13 :  what  is  the  third?  Ans.  25. 


') 


Art.  56.    PEOMISCTJOUS  examples. 

1.  In  4  bags  are  $500:  in  the  first,  96;  the  2d,  120; 
the  3d,  55  :  what  sum  in  the  4th  bag  ?  Ans.  8229. 

2.  Four  men  paid  $1265  for  land;  the  first  paid  8243; 
the  2d,  861  more  than  the  first;  the  3d,  879  less  than  the 
2d :  how  much  did  the  4th  man  pay?  A71S.  $493. 

3.  I  have  5  apple-trees ;  the  first  bears  157  apples ;  2d, 
264  ;  3d,  305  ;  4th,  97 ;  5th,  123  :  I  sell  428,  and  186 
are  stolen  :  how  many  apples  are  left  ?  Ans.  332. 


Review. — 54.  "What  is  found  by  multiplication?  Give  an  example. 
When  the  divisor  and  quotient  are  given,  how  is  the  dividend  found  ? 

55.  What  is  found  by  division  ?  What  does  it  enable  us  to  do  ?  Give 
examples.  If  the  dividend  and  quotient  are  given,  how  is  the  divisor 
found  ?  Having  the  product  of  three  numbers,  and  two  of  them  given, 
how  is  the  other  found  ? 


64  RAY'S    PRACTICAL   ARITHMETIC. 

4.  In  an  army  of  57068  men,  9503  are  killed  ;  586  join 
the  enemy  ;  4794  are  prisoners  ;  1234  die  of  wounds  ;  850 
are  drowned  :  how  many  return  ?  Ans.  40101. 

5.  On  the  first  of  the  year  a  man  is  worth  ^123078  ; 
during  the  year  he  gains  ^8706  ;  in  January  he  spends 
$237,  in  February  8301  ;  in  each  of  the  remaining  ten 
months,  he  spends  as  much  as  in  the  first  two  :  how  much 
had  he  at  the  end  of  the  year?  Ans.  8125866. 

6.  In  a  building  there  are  72  rooms  ;  in  each  room  4 
windows,  and  in  each  window  24  lights  :  how  many  lights 
are  there  in  the  house?  Ans.  6912. 

7.  A  merchant  has  9  pieces  of  cloth,  of  73  yards  each  : 
and  12  pieces,  of  88  yards  each  :  how  many  yards  in  all  ? 

Alls.  1713  yards. 

8.  I  spend  99  cents  a  day  :  how  many  cents  will  I  spend 
in  49  years,  of  365  days  each?  Ans.  1770615. 

9.  An  Encyclopedia  consists  of  39  volumes ;  each  vol- 
ume has  774  pages  of  two  columns  each  ;  each  column  67 
lines ;  each  line  10  words  ;  and  every  10  words  47  letters : 
how  many  pages,  lines,  words,  and  letters,  in  the  work  ? 

.       I        30186  pages,       4044924  lines, 
^"^-  I  40449240  words,  190111428  letters. 

10.  The  Bible  has  31173  verses  :  in  how  many  days  can 
I  read  it,  reading  86  verses  a  day?        Ans.  362|J  days. 

11.  I  bought  28  horses  for  $1400;  3  died:  for  how 
much  each  must  I  sell  the  rest,  to  incur  no  loss  ?  Ans.  $56. 

12.  How  many  times  can  I  fill  a  15  gallon  cask,  from 
5  hogsheads  of  63  gallons  each  ?  Ans.  21  times. 

13.  A  certain  dividend  is  73900  ;  the  quotient  214  ;  the 
remainder  70  :  what  is  the  divisor  ?  Ans.  345. 

14.  Multiply  the  sum  of  148  and  56  by  their  diiference  ; 
divide  the  product  by  23.  Ans.  816. 

15.  How  much  cloth,  at  $6  a  yard,  will  pay  for  8  horses 
at  $60  each,  and  14  cows  at  $15  each  ?    Ans.  115  yards. 

16.  A  cistern  of  360  gallons,  has  2  pipes;  one  will  Jill 
it  in  15  hours,  and  the  other  empty  it  in  20  hours.  If 
both  pipes  are  left  open,  how  many  hours  will  the  cistern 
be  in  filling?  Ans.  60  hours. 


REVIEW    OF    PRINCIPLES.  65 

17.  Two  men  paid  $6000  for  a  farm  ;  one  man  took  70 
acres  at  §30  an  acre,  the  other  the  remainder,  at  $25  an 
acre  :  how  many  acres  in  all  ?  Ans.  226  acres. 

Suggestion. — In  the  four  following  examples,  obtain  the  re- 
quired number  by  reversing  the  operations. 

18.  What  is  the  number,  from  which,  if  125  be  sub- 
tracted, the  remainder  will  be  222  ?  Ans.  347. 

19.  What  is  the  number,  to  which,  if  135  be  added,  the 
sum  will  be  500  ?  Ans.  365. 

20.  Find  a  number,  from  which,  if  65  be  subtracted, 
and  the  remainder  divided  by  15,  the  quotient  will 
be  45  ?  Ans.  740. 

21.  What  is  the  number,  to  which  if  15  be  added,  the 
sum  multiplied  by  9,  and  11  taken  from  the  product,  the 
remainder  will  be  340  ?  Ans.  24. 

22.  If  98  be  subtracted  from  the  difference  of  two 
numbers,  27  will  remain  ',  246  is  the  less  number :  what 
is  the  greater?  Ans.  371. 

GExiERAL   PRINCIPLES    OF    DIVISIOIT. 

Art.  57.  The  value  of  the  quotient  depends  on  the 
relative  values  of  divisor  and  dividend.  These  may  be 
changed  by  Multiplication  and  Division,  thus  : 

1st.  The  Dividend  may  he  multiplied^  or  the  Divisor  divided. 

2d.  The  Dividend  may  he  divided^  or  the  Divisor  multiplied. 

3d.  Both  Dividend  and  Divisor  may  he  multiplied,  or  both 
divided,  at  the  same  time. 

ILLUSTRATIONS. — Let  24  be  a  dividend,  and  6  the  divi- 


sor; the  quotient  is  4:   (24-^6  =  4). 

If  the  dividend  (24),  be  multiplied  by  2,  the  quotient  will 
be  multiplied  by  2:  for,  24X2  =  48;  and  48 -7- 6  =  8,  which 
is  the  former  quotient  (4),  multiplied  by  2. 

Now,  if  the  divisor  (G),  be  divided  by  2,  the  quotient  will  be 
multiplied  by  2;  for,  6-i-2  =  3;  and  24-r-3  =  8,  which  is 
the  former  quotient  (4),  multiplied  by  2.     Hence, 

Principle  1. — If  the  dividend  he  multiplied,  or  the  divisor 
be  divided,  the  quotient  will  be  multiplied. 
3d  Bk.  5 


66  RAY'S   PRACTICAL    ARITHMETIC. 

Art.  58.  Take  the  same  Example,  2-4-^6  =  4. 

If  the  dividend  (24),  be  divided  by  2,  the  quotient  will  be 
divided  by  2  :  for,  24  -j-  2  =  12 ;  and  12  -r-  6  =  2,  which  is 
the  former  quotient  (4),  divided  by  2. 

And,  if  the  divisor  (6),  be  multiplied  by  2,  the  quotient  will 
be  divided  by  2;  for,  6X2=12;  and  24 -r- 12  =  2,  which 
is  the  former  quotient  (4),  divided  by  2.     Hence, 

Principle  II. — If  the  dividend  be  divided,  or  the  divisor  he 
multiplied,  the  quotient  will  be  divided. 

Art.  59.  Take  the  same  Example,  24-^6  =  4. 

If  the  dividend  (24),  and  divisor  (6),  be  multiplied  by  2,  the 
quotient  will  not  be  changed;  for,  24X2=48;  and  6X2  =  12; 
48-7-12^=^4;  the  former  quotient  (4),  unchanged. 

And  if  the  dividend  (24),  and  divisor  (6),  be  divided  by  2, 
the  quotient  will  not  be  changed;  for,  24 -r- 2  =  12;  and 
6-7-2^=3;    12-7-3=::4;  the  former  quotient  (4),  unchanged. 

Hence, 

pRixciPLE  III. — If  both  dividend  and  divisor  be  multiplied  or 
divided  by  the  same  number,  the  quotient  will  not  be  changed. 

Art.  60.  If  a  number  he  multiplied,  and  the  p)roduct 
divided  by  the  same  number,  the  quotient  toill  he  the  original 
number. 

For,  24  X  2  =48 ;  and  48  -r-  2  =  24,  the  original  number, 
on  the  principle,  that  if  the  product  be  divided  by  the  multi- 
plier, the  quotient  will  be  the  multiplicand.     Also, 

If  a  number  he  divided,  and  the  quotient  multiplied  hi/ the 
same  number,  the  product  will  he  the  original  number. 

For,  24 -h  2  =  12;  and  12X2  =  24,  the  original  number, 
on  the  principle,  that  if  the  quotient  be  multiplied  by  the 
divisor,  the  product  will  be  the  dividend. 

Hence,  the  operations  of  multiplication  and  division  by  the 
same  number,  destroy  (cancel)  each  other. 

Eeview. — 57.  On  what  docs  the  value  of  the  quotient  depend  ?  How 
may  the  divisor  and  dividend  be  changed  ?  If  the  dividend  be  multiplied, 
what  effect  on  the  quotient  ?     If  the  divisor  be  divided  ? 

58.  If  the  dividend  bo  divided,  what  effect  on  the  quotient?  If  the 
divisor  be  multiiiliod? 

5U.  If  both  divisor  and  dividend  be  multiplied  by  the  same  number, 
what  effect  on  the  quotient  ?     If  both  be  divided  by  the  same  number  ? 


VI.   CANCELLATION. 

Most  Teachers  defer  this  subject  until  after  Factoring. 

1.  I  bought  3  oranges  at  10  cents  each,  and  paid  for 
them  with  pears  at  3  cts.  each  :  how  many  pears  did  I  give  ? 

Solution. — Ten  cents  multiplied  by  3,  operation. 

give  30  cents,  the  cost  of  the  oranges.  1  0 

Then  it  will  take  as  many  pears,  as 3  cents  3 
are  contained  times  in  30  cents;   that  is, 

30-^3  =  10,  the  number  of  pears.  S)30 

Here,  10  is  multiplied  by  3  and  the  pro-  ^^^^    i  Q  pears, 
duct  divided  by  3;  but  a  number  is  not  changed 

by  multiplying  it,  and  then  dividing  the  product  by  the  same  number, 
(Art.  60) ;  hence,  multiplying  by  3,  and  then  dividing  by  3,  may 
be  omitted,  and  10  taken  as  the  result ;    hence, 

Art.  61   .  When  a  number  is  to  be  mul-  operation. 

tiplied  and  then  divided  by  the  same  num-  1  0  X   ^ 

ber,  both  operations  may  be  omitted,  and    ~       =10. 

a  Hne  drawn  across  the  common  multiplier  P 
and  divisor,  as  in  the  margin. 

Eem, — In  the  above  example,  10  and  3  form  the  dividend,  and  3  the 
di\asor.  In  arranging  the  numbers,  place  the  dividend  above  a  horizontal 
line,  and  the  divisor  below  it. 

2.  How  many  barrels  of  molasses  at  $13  a  barrel,  will 
pay  for  13  barrels  of  flour  at  $4  a  barrel  ?         Ans.  4. 

3.  If  I  buy  41  cows  at  Sll  each,  and  pay  in  horses  at 
$41  each,  how  many  horses  are  required?         Ans.  11. 

4.  If  I  buy  10   lemons   at   3   cents  each,   and   pay  in 
oranges  at  5  cts.  each,  how  many  oranges  will  1  give  ? 

Solution. — Ten  times   3   cents  are    30  cents,      operation. 
the  cost  of  the  lemons  :  30  cts.  divided  by  5  cents,  3 

equal  6,  the  number  of  oranges.  1 0 

But,  as  10  is  a  composite  number,  whose  factors        k Von^ 

are  5  and  2,  (5X2  =  10),  indicate  the  operation  I 

as  in  the  margin  on  the  left.         Ans.  6. 
0X2  X  3_  Since  5,  2,  and  3  are  to  be 

Z  multiplied  together,  and  their  product  divided 

^  by  5,   omit  5  both    as   multiplier   and   divisor 

(Art.  60),  draw  a  line  across  it,  and  only  multiply  2  by  3. 


68  RAY'S   PRACTICAL  ARITHMETIC. 

5.  Multiply  17  by  18,  and  divide  the  product  by  6. 

Solution. — Instead  of  mul-  ofekation. 

tiplying  17  by  18,  and  dividing  Dividend,  17X^X3 

the    product    by    6,    separate  ^51. 

18  into   the  factors  6  and  3,  Divisor,  ^ 

(6X3  =  18),  then  cross  the  factor  6,  which  is  common  to  both  mul 

tiplier  and  divisor:  after  which,  multiply  17  by  3. 

6.  In  15  times    8,  how  many  times  4?  Ans.  30. 

7.  In  24  times    4,  how  many  times  8  ?  Ans.  12. 

8.  In  37  times  15,  how  many  times  5?  Ans.  111. 

9.  Multiply  36  by  40,  and  divide  the  product  by  30 
multiplied  by  8. 

OPERATION, 

Dividend,  36X4O_0X3X2X0X^X2_ 
Divisor,       30X8  ^X^X,4X^ 

Solution. — Indicate  the  operation  to  be  performed,  then  resolve 
(separate)  the  numbers  into  factors.  Now  the  quotient  will  not  be 
changed  (Art.  59)  by  dividing  both  divisor  and  dividend  by  the 
same  number,  which  is  done  by  erasing  the  same  factors  in  both. 

Second  Sol. — Indicate  the  oper-  operation. 

ation  as  in  the  margin.  Q  K 

As  8  is  a  factor  of  40,  divide  40       Dividend,  ^  ^  X  ^  0 

by  8,  and  write  the  quotient  5  over  =o. 

40;  cross  out  (cancel)  40  and  8.  Divisor,      ^  0X  g 

Then,  as  36  and  30  have  a  com-  y 

mon  factor  6,  divide  each  by  6,  and  write  the  quotients  6  and  5  as  in 
the  operation :  cancel  36  and  30.  Next,  canceling  the  common  factor 
6  in  botli  dividend  and  divisor,  the  result  is  6,  as  before. 

10.  In  36  times  5,  how  many  times  15?  A71S.  12. 

11.  la  14  times  9,  how  many  times  6?  Ans.  21. 

Art.  61  ^  The  process  of  shortening  the  operations  of 
arithmetic,  by  omitting  equal   factors  from  the  dividend 

Review. — GO.  If  a  number  bo  multiplied,  and  the  product  divided  by 
the  same  number,  what  will  bo  the  quotient?  On  what  principle?  If  a 
number  bo  divided  and  the  quotient  multiplied  by  the  same  number,  what 
will  the  product  bo?     On  what  principle? 


CANCELLATION.  69 

and  divisor,  is  termed  Cancellation.     It  depends   on  tlie 
principle  explained  in  Arts,  58  and  59. 

Note. — To  cancel  is  to  suppress  or  erase.     When  the  same  factor  is 
omitted  in  both  dividend  and  divisor,  it  is  said  to  be  canceled. 

RULE   FOR   CxlNCELLATION. 

When  there  are  common  factors  in  a  dividend  and  its  divisor., 
shorten  the  operation  by  canceling  all  the  factors  common  to  both : 
proceed  with  the  remaining  factors  as  the  question  may  require. 

Rem. — 1.  Canceling  is  merely  dividing  both  dividend  and  divisor  by  the 
same  number,  which  (Art.  59)  does  not  alter  the  quotient. 

2.  The  pupil  should  observe  that  one  factor  in  the  dividend  will  cancel 
only  one  equal  factor  in  the  divisor. 

3.  Some  prefer  to  place  the  numbers  forming  the  dividend  on  the  right 
of  a  vertical  line,  and  those  forming  the  divisor  on  the  left. 

1.  Multiply  42,  25,  and  18,  together,  and  divide  the 
product  by  21X15.  Ans.  60. 

2.  I  sold  23  sheep  at  $10  each,  and  was  paid  in  hogs  at 
$5  each  :  how  many  did  I  receive  ?  Ans.  46. 

3.  How  many  yards  of  flannel  at  35  cents  a  yard,  will 
pay  for  15  yards  of  calico  at  14  cts.  ?  Ans.  6  yards. 

4.  What  is  the  quotient  of  21X11X6X26,  divided  by 
13X3X14X2?  An^.  33. 

5.  The  factors  of  a  dividend  are  21,  15,  33,  8,  14, 
and  17;  the  divisors,  20,  34,  22,  and  27:  required  the 
quotient.  Ans.  49. 

6.  I  bought  21  kegs  of  nails  of  95  pounds  each,  at  6 
cents  a  pound ;  paid  for  them  with  pieces  of  muslin  of  35 
yards  each,  at  9  cents  a  yard ;  how  many  pieces  of  muslin 
did  I  give  ?  Ans.  38. 

Note. — Other  applications  of  Cancellation  will  be  found  in  Fractions, 
Proportion,  &c.  The  pupil  will  apply  it  more  readily,  when  acquainted 
with  Factoring. 


Review. — 61*^.  Rem.  How  are  numbers  arranged  for  cancellation  ? 
610,  What  is  cancellation?  Upon  what  principle  does  it  depend? 
What  does  cancel  mean  ?  "When  is  a  factor  canceled  ?  What  is  the  rule  ? 
Rem.  What  is  canceling  ? 


70  RAY'S    PRACTICAL   ARITHMETIC. 


VII.    COMPOUND    NUMBERS. 

To  Teachers. — "While  placing  Fractions  immediately  after  Simple 
Whole  Numbers  is  philosophical,  and  appropriate  in  a  Higher  Arithmetic 
for  Advanced  pupils,  the  experience  c.f  the  author  convinces  him  that,  in  a 
book  for  Yming  learners,  Compound  Numbers  should  be  introduced  here, 
instead  of  after  Fractions,  as  is  done  by  some  authors.     His  reasons  arc, 

1st.  The  operations  of  Addition,  Subtraction,  Multiplication,  and  Divi- 
sion of  Compound  Nunibers,  are  analogous  to  the  same  operations  in  Simple 
Ifumbers,  and  serve  to  illustrate  the  principles  of  the  fundamental  rules. 
The  principle  oi  Notation  is  the  same  in  each. 

2d.  The  subject  of  Fractions  is  important  and  difficult.  Before  studying 
it,  most  pupils  require  more  mental  discipline  than  is  demanded  in  the  ele- 
mentary rules.     This  is  acquired  by  the  study  of  Compound  Numbers. 

3d.  The  general  principles  involved  in  their  study,  do  not  require  a 
knowledge  of  fractions.  The  Examples  involving  fractions  are  few,  and  are 
introduced,  (as  they  should  be,)  with  other  exercises  in  that  subject. 

7^^^  Teachers  who  prefer  it,  can  direct  their  pupils  to  defer  Compound 
Numbers  until  they  have  studied  Fractions  to  page  169. 

DEFINITIONS. 

Art.  62.  When  two  numbers  have  the  %aine  unit^  they 
are  of  the  same  kind  or  denomination  :  thus,  3  dollars,  5 
dollars,  are  of  the  same  denomination ;  both  dollars. 

When  they  have  different  units,  they  are  of  different 
denominations  :  thus,  3  dollars,  and  5  cents,  are  of  different 
denominations  ;   dollars  and  cents. 

Art.  63.  A  simple  number  denotes  things  of  the  same 
unit  value:  thus,  3  yards,  2  dollars,  5  pints,  are  each  simple 
numbers.     All  abstract  numbers  are  simple. 

Art.  64.  A  compound  number  is  two  or  more  numbers 
of  different  unit  values  used  to  express  one  quantity :  thus, 
3  dollars  5  cents,  2  feet  3  inches,  are  each  compound. 

Rem. — 1.  In  Compound  Numbers,  denomination  or  order,  denotes  the 

Review. — ()2.  When  are  two  numbers  of  the  samo  denomination  ?  Give 
an  example.     When  of  different  denominations?     Give  an  example. 

63.  What  does  a  Simple  Number  denote?  Give  an  example.  64.  What 
is  a  Compound  Number?     Give  an  example. 


COMPOUND  NUMBERS.     U.  S.  MONEY.  71 

name  of  the  unit  considered.     Thus,  dollar  and  cent  are  denominations  of 
money ;   foot  and  inch,  of  length  ;    povnd  and  ounce,  of  weight. 

2.  Compound  Numbers  are  analogous  to  Simple  Numbers  in  this  particu- 
lar ;  a  certain  number  of  units  of  each  order  is  collected  into  a  group,  and 
forms  a  unit  of  a  higher  order  or  denomination.     But, 

They  diflfer  in  this  :  that  in  compound  numbers  10  units  of  one  order  do 
not  unifonnly  make  one  of  the  next  higher. 

3.  The  simplest  class  of  Compound  Numbers  is  Federal  money,  because 
we  pass  from  one  denomination  to  another  according  to  the  scale  of  tens. 

FEDERAL  OR  UNITED   STATES   MONEY, 

Art.  65,  Is  tlie  currency  of  the  United  States,  estab- 
lished by  the  Federal  Congress,  in  1786. 

While  U.  S.  money  may  be  treated  decimally,  it  is  a  species  of  Compound 
Numbers,  being  so  regarded  in  ordinary  business  transactions. 

Its  denominations,  or  the  names  of  its  different  orders,  are 
mill,  cent,  dime,  dollar,  eagle. 

Ten  units  of  each  denomination  make  one  unit  of  the  next 
higher  denomination. 

TABLE. 

10  mills,  marked  ra.,  make  1  cent,     marked  ct. 

10  cents      1  dime,      d. 

10  dimes     1  dollar,    $. 

10  dollars 1  eagle,     b. 

Also,     5  cents make  one-half  dime. 

25  cents one-quarter  of  a  dollar. 

50  cents one-half  of  a  dollar. 

75  cents three-quarters  of  a  dollar. 

100  cents one  dollar. 

The  coins  of  the  United  States  are  of  bronze,  nickel, 
silver,  and  gold.     Their  denoniinations  are : 

1st.  Bronze:  cent.    2d.  Nickel :  three-cent  piece,  five-cent  pieca 

3d.   Silver:  dollar,  half  dollar,  quarter  dollar,  dime. 

4th.  Gold:  double  eagle,  eagle,  half  eagle,  three-dollar  piece, 
quarter  eagle,  dollar. 


Review. — 64.  Eem.  1.  What  does  den/rmination  or  order  denote  ?  2.  In 
what  are  Simple  and  C(jmpound  Numbers  analogous  ?  In  what  do  they 
ditfer  ?    3.  What  is  the  simplest  class  of  Compound  Numbers  ? 


I 


72  RAYS   PRACTICAL   ARITHMETIC. 

NOTATION    AND    NUMERATION. 

Art.  66.    Accounts   are    kept    in    dollars,    cents,    and 
mills ;  or  irn  dollars,  cents,  and  parts  of  a  cent. 

Eagles  and  dollars  are  called  dollars ;     dimes  and  cents,  cents. 


bC 

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NUiMERATION    TABLE. 


read  4  cents  and  3  miilr>,  or  43  mill-^. 

read  21  cents  and  4  mills,  or  214  mills. 

read  3  dollars  4  cents  and  5  mills. 

read  76  dollars  and  25  cents, 

read  681  dollars  34  cents  and  5  mills. 

The  second  line  may  also  be  read,  2  dimes,  1  cent,  4  mills; 
the  fourth  line,  7  eagles,  6  dollars,  2  dimes,  5  cents.  This 
method  of  reading  is  not  customary. 

The  third  line  may  also  be  read,  304  cents  and  5  mills,  or 
3045  mills;  the  fourth  line,  7625  cents,  or  76250  mills;  the 
lower  line,  68134  cents  and  5  mills,  or  681345  mills. 

A  period  (.),  is  used  as  a  separating  point,  to  separate  the  cents 
and  dollars.     Some  use  the  comma. 

Thus,  2  dollars,  2  dimes,  and  2  cents,  or  2  dollars  and  22  cents, 
are  written,  $2.22. 

Art.  67.  The  Table  shows  that  cents  occupy  the  first 
two  places  to  the  right  of  dollars,  and  mills  the  place  to 
the  right  of  cents,  the  third  from  dollars.     Hence  the 

Rule  for  Nmneration. — Read  the  number  to  the  left  of  the 
period  as  dollars,  and  the  first  two  fgures  on  the  right  of  the 
period  as  cents ;  and  if  there  he  a  third  figure,  as  mills. 

Revikw. — 65.  What  are  the  denominations  of  United  States  money? 
IIow  m:iny  units  of  cither  denomination  make  a  unit  of  the  next  higher  ? 
Repeat  the  table.  IIdw  many  cents  in  a  half  dima  ?  In  a  quarlcr  dollar? 
In  a  half  dollar  ?  In  a  dollar  ?  Of  what  are  the  coins  of  the  United  States  ? 
Which  are  copper?     Which  silver?    Which  gold? 


COMPOUND  NUMBERS.      U.   S.   MONEY.  73 

EXAMPLES   TO   BE   COPIED   AND   THEN   READ. 

$18.62  5  $  70.01  5  $6.12  $  29.00 

$20.32  4  $100.28  3  S3. 06  $100.03 

$79.05  $150.00  2  $4.31  $  20.05 

846.00  3  $100.00  3  $5.43  $  40.00  7 

Art.  68.  RULE   FOR   NOTATION. 

Write  the  dollars  as  in  whole  numbers  ;  place  a  period  on  the 
right  of  dollars,  next  to  this  write  the  cents,  then  the  mills. 

If  the  cents  are  less  than  ten,  place  a  cipher  next  to  the  dollars; 
if  there  are  no  cents,  put  two  ciphers  in  the  place  of  cents. 

EXAMPLES   TO   BE   WRITTEN. 

1.  Twelve  dollars,  seventeen  cents,  eight  mills,  $12.17  8 

2.  Six  dollars,  six  cents,  six  mills %  6.06  6 

3.  Seven  dollars,  seven  mills %  7.00  7 

4.  Forty  dollars,  fifty-three  cents,  four  mills.  .     $40.53  4 

5.  Two  dollars,  three  cents $  2.03 

6.  Twenty  dollars,  two  cents,  two  mills.     .     .     $20.02  2 

7.  One  hundred  dollars,  ten  cents $100.10 

8.  Two  hundred  dollars,  two  cents $200.02 

9.  Four  hundred  dollars,  one  cent,  eight  mills.  $400.01  8 

EEDUCTION    OF    U.   S.    MONEY. 

Art.  69.  Reduction  consists  in  changing  the  denomi- 
nations or  orders,  without  altering  the  value. 

Ist.  Reduction  Descending  is  changing  numbers  from  a 
higher  to  a  lower  denomination ;     as,  from  dollars  to  cents. 

2d.  Reduction  Ascending  is  changing  numbers  from  a  lower 
to  a  higher  denomination  ;     as,  from  cents  to  dollars. 

Review. — 6(5.  In  what  denominations  are  accounts  kept?  "What  are 
eagles  and  dollars  together  called  ?  What  dimes  and  cents  ?  How  are 
dollars  and  cents  separated  ?  67.  "What  places  do  cents  occupy  ?  "What 
place  mills  ?     "What  is  the  Rule  for  Numeration  ? 

68.  How  write  any  sum  of  U.  S,  money,  Rule  ?  69.  In  what  does  Reduc- 
tion consist?    "What  is  Reduction  descending?    "What  ascending? 


74  RAY'S    PRACTICAL    ARITHMETIC. 

Art.  70.  As  there  are  10  mills  in  1  cent,  in  any  num- 
ber of  cents  there  are  10  times  as  many  mills  as  cents. 

Therefore,  to  reduce  cents  to  mills,  multiply  by  10;  that  is, 
annex  one  cipher. 

Hence,  conversely,  to  reduce  mills  to  cents,  divide  by  10; 
that  is,  cut  off  one  figure  on  the  right. 

As  there- are  100  cents  in  1  dollar,  in  any  number  of 
dollars  there  are  100  times  as  many  cents  as  dollars. 

Therefore,  to  reduce  dollars  to  cents,  multiply  by  100 ;  that 
is,  annex  two  ciphers. 

Hence,  conversely,  to  reduce  cents  to  dollars,  divide  by  100; 
that  is,  cut  off  two  figures  on  the  right 

As  there  are  1000  mills  in  a  dollar,  in  any  number  of 
dollars  there  are  1000  times  as  many  mills  as  dollars. 

Therefore,  to  reduce  dollars  to  mills,  multiply  by  1000 ;  that 
is,  annex  three  ciphers. 

Hence,  conversely,  to  reduce  mills  to  dollars,  divide  by  1000; 
that  is,  cut  off  three  figures  on  the  right. 

Art.  71.  As  the  operations  of  reduction  consist  in 
multiplying  or  dividing  by  10,  100,  or  1000,  the  work 
can  be  shortened  by  simply  moving  the  point. 

Illustrations. — In  Multiplying^  move  the  point  as  many  places 
to  tlje  right^  as  there  are  ciphers  in  the  multiplier. 

Thus,  $2.50  =  250  cents;     $2.50  5  =  2505  mills. 

In  Dividing^  move  the  point  as  many  places  to  the  left,  as  there 
are  ciphers  in  the  divisor. 

Thus,    275  cents  =  $2.75;     4285  mills  =  $4.28  5 

1.  Reduce  17  cts.  to  mills.  Ans.  170  m. 

2.  lleduee  28  cts.  to  mills.  Ans.  280  m. 


Review. — 70.  Ilnw  are  cents  reduced  to  mills  ?  "Why  ?  Mills  to  cents  ? 
Dollard  to  cents  ?  Why  ?  Cents  to  dollars  ?  Dollars  to  mills  ?  Why  ? 
Mills  to  dollars?  71.  In  what  do  the  operations  of  reduction  consist? 
How  can  the  work  bo  shortened  ?     Illustrate. 


COiMPOUND  NUMBERS.      U.   S.   MONEY. 


75 


3.  Reduce 

4.  Keduce 
Reduce 
Reduce 
Reduce 
Reduce 
Reduce 

10.  Reduce 

11.  Reduce 
Reduce 
Reduce 
Reduce 

15.  Reduce 

16.  Reduce 

17.  Reduce 

18.  Reduce 


5. 
G. 

7. 
8. 
9. 


12. 
13. 
14. 


43  cts.  and  6  m.  to  mills 
70  cts.  and  6  m.  to  mills 
106  m.  to  cents. 
490  mills  to  cents. 
9  dollars  to  cents. 
14  dollars  to  cents. 
104  dollars  to  cents. 
^60  and  13  cts.  to  cents. 
^0  and  5  cts.  to  cents. 
375  cts.  to  dollars. 
9004  cts.  to  dollars. 
4  dollars  to  mills. 
814  and  2  cts.  to  mills. 
2465  mills  to  dollars. 
3007  mills  to  dollars. 
3187  cents  to  dollars. 


Ans.  436  m. 

Ana.  706  m. 

Ans.  10  cts.  6  m. 

Ans.  49  cts. 

A?is.  900  cts. 

Ans.  1400  cts. 

Ans.  10400  cts. 

Ans.  6013  cts. 

Ans.  4005  cts. 

Ans.  83.75 

Ans.  $90.04 

Ans.  4000  m. 

Ans.  14020  m. 

Ans.  $2.46  5 

Ans.  $3.00  7 

Ans.  $31.87 


ADDITION    OF    TJ.   S.    MONEY. 

Art.  72.  1.  Add  together  4  dollars,  12  cents,  5  mills; 
7  dollars,  6  cents,  2  mills;  20  dollars  43  cents;  10  dol- 
lars, 5  mills ;  and  16  dollars,  87  cents,  5  mills. 


Rule — 1.  Write  the  numbers  to  he  added, 
units  of'  the  same  denominations  under  each 
other;  dollars  under  dollars,  cents  under 
cents,  mills  under  mills,  because  only  num- 
bers of  the  same  denomination  can  be  added. 

2.  Add  as  in  Addition  of  Simple  Num- 
bers, and  place  the  separating  point  directly 
under  the  sejjarating  points  above. 


OPERATION. 

$ 

cts. 

rn. 

4 

.12 

5 

7 

.06 

2 

20 

43 

0 

10. 

00 

5 

16 

87 

5 

$58.49  7 


Proof. — The  same  as  ia  Addition  of  Simple  Numbers. 


Eeview. — 72.  How  write  numbers  in  addition  of  U.  S.  money?     Why? 
How  id  the  addition  performed  ?    Why  ?     Where  place  the  point  ? 


76  RAY'S   PRACTICAL   ARITHMETIC. 

2.  Wliat  is  the  sum  of  17  dollars,  15  cents  ;  23  dollars, 
43  cents  ;  7  dollars,  19  cents  ;  8  dollars,  37  cents ;  and 
12  dollars,  31  cents  ?  Ans.  $68.45 

3.  Add  18  dollars,  4  cents,  1  mill ;  16  dollars,  31  cents, 
7  mills ;  100  dollars,  50  cents,  3  mills  ;  and  87  dollars, 
33  cents,  8  mills.  Ans.  §222.19  9 

4.  "William  had  the  following  bills  for  collection  : 
$43.75;  $29.18;  $17.63;  $268.95;  and  $718.07:  how 
much  was  to  be  collected?  Ans.  $1077.58 

5.  Bought  a  gig  for  $200;  a  watch  for  $43.87  5;  a 
suit  of  clothes  for  $56.93  7  ;  hat  for  $8.50 ;  and  a  whip 
for  $2.31  3:  what  was  the  amount?     Ans.  $311.62  5 

6.  A  person  has  due  him,  five  hundred  and  four 
dollars,  six  cents,  three  mills;  $420,  19  cents,  7  mills; 
one  hundred  and  five  dollars,  fifty  cents ;  $304  and  5 
mills ;  $888,  forty-five  cents,  five  mills  :  how  much  is  due 
to  him?  Ans.  $2222.22 

7.  Add  five  dollars,  seven  cents ;  thirty  dollars,  twenty 
cents,  three  mills ;  one  hundred  dollars,  five  mills ;  sixty 
dollars,  two  cents ;  seven  hundred  dollars,  one  cent, 
one  mill;  $1000.10;  forty  dollars,  four  mills;  and 
$64.58  7  Ans.  $2000. 

SUBTRACTION    OF    U.  S.    MONEY. 

Art.  73.  From  one  hundred  dollars,  five  cents,  three 
mills,  take  $80,  20  cents,  7  mills. 


Hule. — Place  the  less  niimher  under 


OPERATION. 

$      cts.  m. 
the  greater^  dollars  under  dollars^  cents  inn    af;  Q 

under  cents,  <fcc.    Subtract  as  in  Simple  Q.f\'  9(\  7 

Numbers,  placing  the  separating  point  ' 

under  the  points  above.  Ans.  $19.84  6 

Proof. — As  in  Subtraction  of  Simple  Numbers. 

2.  From  $29.34  2  take  $17.26  5        Ans.  $12.07  7 

3.  From  $46.28       take  $17.75  Ans.  $28.53 

4.  From  $20.05       take  $  5.50  Ans.  $14.55 


COMPOUND   NUMBERS.      U.   S.    MONEY.  77 

5.  From  $3,  take  3  cts.  .     .     .     Ans.    $2.97 

6.  From  $10,  take  1  mill.      .     .     .     Ans.     $9.99  9 

7.  From  $50,  take  50  cts.,  5  mills.      Ans.  $49.49  5 

8.  From  one  thousand  dollars,  take  one  dollar,  one 
cent,  and  one  mill.                                    Ans.  $998.98  9 

9.  B  owes  1000  dollars,  43  cents,  5  mills;  if  lie  pay 
nine  hundred  dollars,  sixty-eight  cents,  seven  mills,  how 
much  will  he  still  owe?  Ans.  $99.74  8 

MULTIPLICATION    OF    U.    S.    MONEY. 

Art.  74.  1.  What  will  13  cows  cost,  at  17  dollars,  12 
cents,  5  mills  each  ?  ;. 

Solution. — Consider    the   cost,  $17.12  5,  as  operation. 

reduced  to  its  lowest,  denomination,  viz.:   17125  $17.12  5 

mills.    Then,  since  13  cows  will  cost  13  times  as  13 

much  as  1,  multiply  the  cost  of  one,  by  13,  which  ~     ;      ~" 

gives  for  the  cost  of  13  cows,  222025  mills,  the  51375 

1  7  1  '^  5 

product  being  of  the  same  denomination  as  the  * 

multiplicand,  Art.  30.     Finally,  reduce  the  mills         $222.62  5 
to  dollars,  Art.  70.     Hence,  the 

Rllle. — Multiply  as  in  Simple  Numbers  ;  the  product  will  be 
the  answer  in  the  lowest  denomination  of  the  multiplicand^ 
which  may  then  be  reduced  to  dollars  by  pointing. 

Proof. — As  in  Multiplication  of  Simple  Numbers. 

2.  Multiply  $7,  83 cts.   by  8.  Ans.      $62.64 

3.  Multiply  $12,  9  cts.,  3  m.   by  9.  Ans.    $108.83  7 

4.  Multiply  $23,  let.,  8m.   by  16.  Ans.    $368.28  8 

5.  Multiply  $35,  14  cts.   by  53.  Ans.  $1862.42 

6.  Multiply  $125,  2  cts.   by  62.  Ans.  $7751.24 

7.  Multiply  $40,  4  cts.   by  102.  Ans,  $4084.08 

8.  Multiply  12  cts.,  5  m.  by  17.  Ans.        $2.12  5 

9.  Multiply  $3,  28  cts.  by  38.  Ans.    $124.64 

Review. — 73.  How  are  numbers  written  in  subtraction  ?  "Why  ?  How 
is  the  subtraction  performed?  Where  is  the  separating  point  placed? 
74.  How  is  multiplication  performed  ?    How  is  the  product  pointed  ? 


78  RAY'S    PRACTICAL    ARITHMETIC. 

10.  What  cost  338  barrels  of  cider,  at  1  dollar,  6  cents 
a  barrel?  ^7is.  $358.28 

11.  Sold  38  cords  of  wood,  at  5  dollars,  75  cts.  a  cord : 
to  what  did  it  amount?  Ans.  $218.50 

12.  At  7  cts.  a  pound,  what  cost  465  pounds  of  sugar? 

Ans.  $32.55 
Note. — Instead  of  multiplying  7  cents  by  465,  multiply  465  by  7,  which 
gives  the  same  product,  Art.  30.     But,  in  fixing  the  denomination  of  the 
product,  remember  that  7  cents  is  the  true  multiplicand. 

13.  What  cost  89  yards  of  sheeting,  at  34  cts.  a  yard  ? 

Alls.  $30.26 

14.  What  will  24  yards  of  cloth  cost,  at  5  dollars,  67 
cents  a  yard?  Ans.  $136.08 

15.  I  have  169  sheep,  valued  at  $2.69  each:  what  is 
the  value  of  the  whole?  Ans.  $454.61 

16.  If  I  sell  691  bushels  of  wheat  at  1  dollar,  25  cts.  a 
bushel,  what  will  it  amount  to  ?  Ans.  $863.75 

17.  I  sold  73  hogsheads  of  molasses,  of  63  gallons  each, 
at  55  cts.  a  gallon  :  what  is  the  sum?       Ans.  $2529.45 

18.  What  cost  4  barrels  of  sugar,  of  281  pounds  each, 
at  6  cents,  5  mills  a  pound?  Ans.  $73.06 

19.  Bought  35  bolts  of  tape,  of  10  yards  each,  at  1  cent 
a  yard  :  what  did  it  cost?  Ans.  $3.50 

20.  If  I  earn  13  cts.  an  hour,  and  work  11  hours  a  day, 
how  much  will  I  earn  in  312  days?  Ans.  $446.16 

21.  I  sold  18  bags  of  wheat,  of  3  bushels  each,  at  $1.25 
a  bushel :  what  is  the  amount?  Aus.  $67.50 

22.  What  cost  150  acres  of  land,  at  10  dollars,  1  mill 
per  acre?  ^«s.  $1500.15 

23.  What  cost  17  bags  of  coffee,  of  51  pounds  each,  at 
14  cents,  7  mills  per  pound?  Ajis.  $127.44  9 

Art.  75.  DIVISION  of  u.  s.  money. 
The  object  in  Division  of  United  States  money  is, 

Ist.  To  find  how  many  times  one  sum  of  money  is  contained 
in  another  of  the  same  order  or  denomi)iaiion.  Or,  2d.  To  divide 
a  sum  of  money  into  a  given  number  of  equal  parts,  Art.  41. 


COMPOUNI^   NUMBERS.      U.   S.   MONEY.  79 

1.  How  much  cloth  at  7  cts.  a  yard,  will  $1.75  buy? 

Anal. — As  1  yai^d  costs  7  cts.,  there  will  be  as  many  yards 
as  7  cts.  are  contained  times  in  175  cts.     175-t-7=25. 

Here,  the  divisor  and  dividend  are  the  same  denomination,  cents; 
and  the  quotient,  how  many  yards,  is  an  abstract  number. 

2.  Divide  G5  dollars  equally  among  8  persons. 

Solution. — In    this    case    it    is   required    to  operation. 

divide   $65  into  8  equal  parts,  that  is,  to  find  $    cts.  ra. 

one-eighth  of  it.    One-eighth  of  $65  is  $8,  with  a       8)65.00   0 

remainder  $1  =  100  cts.    One-eighth  of  100  cts.  

is  12  cts.,  with  a  remainder  of  4  cts.  which  equals  8 . 1  -,   5 

Mills.  ^^  mills.    One-eighth  of  40  mills  is  5  mills;  hence 

8)65  000         one-eighth  of  $65  is  $8.12  5 

~     ^_  The  operation  may  be  performed  by  reducing  the 

"  ■'■  '^^         dollars  to  mills,  then  dividing  by  8,  and  after  this, 
$8.1  2  5         reducing  the  quotient  to  dollars. 

3.  A  farmer  received  $29.61  cents,  for  23  bushels  of 
wheat :  how  much  was  that  per  bushel  ? 

Solution.— To  divide  $29.61  operation. 

into    23   equal    parts,    annex    a     23)  296  1  0  (1  287  mills. 

cipher,  which  reduces  it  to  29610  23  ^^    ^o   ^  , 

mills ;  then  divide  by  23  as  in  '  ' 

Simple  Numbers. 

The  quotient,  1287,  is  mills, 
(Art.  41)  which  reduce  to  $1.28  7, 
(Art.  70j. 

But,  1287  is  not  the  exact  quo- 
tient, as  there  is  a  remainder  of 
9  mills.  It  is,  however,  less  than 
a  mill  of  being  exact,  which  is  9  Rem. 

sufficiently  accurate  for  business  purposes. 

The  sign  -[-  is  annexed  to  denote  a  remainder. 

Rule  for  Division. — 1.  To  Jind  how  many  times  one  sum 
of  money  is  contained  in  another .^  reduce  both  sums  to  the  same 
denomination,  and  divide  as  in  Simple  Numbers. 


23 

46 

201 

184 

170 
161 

80  RAY'S  PRACTICAL  ARITHMETIC. 

2.  To  divide  a  sum  of  money  into  any  number  of  equal  parts, 
reduce  the  sum  to  mills ;  divide  as  in  Simple  Numbers ;  the 
quotient  will  be  mills,  which  reduce  to  dollars. 

Proof — As  in  Division  of  Simple  Numbers. 

4.  How  many  yards  of  calico,  at  8  cents  a  yard,  can  be 
bought  for  $2.b)0?  Ans.  35  yards. 

5.  How  many  yards  of  ribbon,  at  25  cents  a  yard,  can 
be  purchased  for  $3  ?  Ans.  12  yards. 

6.  At  $8.05  a  barrel,  how  many  barrels  of  flour  will 
$161  purchase  ?  Ans.  20  barrels. 

7.  At  7  cents  5  mills  each,  how  many  oranges  can  be 
bought  for  $1.20?  Ans.  16  oranges. 

8.  At  SI.  12  5  per  bushel,  how  many  bushels  of  wheat 
can  be  purchased  for  $234  ?  Ans.  208  bush. 

9.  If  4  acres  of  land  cost  $92.25,  how  much  is  that 
an  acre?  Ans.  $23.06  2-\- 

10.  Make  an  equal  division  of  $57  and  50  cents  among 
8  persons.  Ans.  $7.18  7-h 

11.  A  man  received  $25  and  76  cts.  for  16  days'  work : 
how  much  was  that  a  day?  Ans.  $1.61 

12.  I  bought  755  bushels  of  apples  for  $328,  42  cts., 
5  mills  :  what  did  they  cost  a  bushel?        Ans.  $0.43  5 

13.  My  salary  is  $800  a  year  :  how  much  is  that  a  day, 
313  working  days  in  the  year?  Ans  $2.55  h-\- 

14.  Divide  ten  thousand  dollars  equally  among  133 
men  :  what  is  each  man's  share?  Ans.  $75.18  7+ 

15.  A  man  purchased  a  farm  of  154  acres,  for  two 
thousand  seven  hundred  and  five  dollars  and  one  cent : 
what  did  it  cost  per  acre?  Ans.  $17.56  5 

16.  I  sold  15  kegs  of  butter,  of  25  pounds  each,  for 
$60  :  how  much  was  that  a  pound?  Ans.  16  cts. 

17.  I  bought  8  barrels  of  sugar,  of  235  pounds  each, 
for  $122.20  :  what  did  1  pound  cost?        Ans.  $0.06  5 

Kkview. — 75.  What  is  the  object  of  division?  How  is  the  number 
of  times  one  sum  of  money  is  contained  in  another  found,  Rule  1  ?  IIow 
is  a  sum  of  money  divided  into  eifual  parts,  Rule  2  ? 


COMPOUND  NUMBERS.     U.   S.   MONEY.  81 

Art.  76. — PEOMISCUOUS     EXAMPLES. 

1.  I  owe  A  $47.50;  B  838.45;  C  815.47;  D 
$19.43  :  what  sum  do  I  owe?  Ans.  8120.85 

2.  A  owes  835.25  ;  B  823.75  ;  C  as  much  as  A  and 
B,  and  81  more  :  what  is  the  amount?  Ans.  8119. 

3.  A  paid  me  818.38;  B  881.62;  C,  twice  as  much 
as  A  and  B  :  how  much  did  I  receive  ?  Ans.  8300. 

4.  I  went  to  market  with  85  ;  I  spent  for  butter  75  cts., 
for  eggs  35  cts.,  for  vegetables  50  cts.,  for  flour  81.50  : 
how  much  money  was  left?  Ans.  81.90 

5.  A  lady  had  820;  she  bought  a  dress  for  88.10, 
shoes  for  81.65,  eight  yards  of  calico  at  75  cts.  a  yard,  and 
a  shawl  for  84  :  what  sum  was  left?  Ans.  25  cts. 

6.  I  get  850  a  month,  and  spend  830.50  of  it:  how 
much  will  I  have  left  in  6  months  ?  Ans.  8117. 

7.  A  farmer  sold  his  marketing  for  821.75:  he  paid 
for  sugar  83.85,  for  tea  81.25,  for  coffee  82.50,  for  spices 
81.50  :  how  much  had  he  left?  Ans.  812.65 

8.  I  owe  A  837.06,  B  8200.85,  C  ^00,  D  8236.75, 
and  E  8124.34;  my  property  is  worth  8889.25:  how 
much  do  I  owe  more  than  I  am  worth?     Ans.  8109.75 

9.  Bought  143  pounds  of  coffee  at  13  cts.  a  pound : 
after  paying  812.60,  what  was  due?  Ans.  85.99 

|p'    10.  A  owed  me  8400  :  he  paid  me  435  bushels  of  corn, 
at  45  cts.  a  bushel :  what  sum  is  due?        Ans.  8204.25 

11.  If  B  spend  65  cts.  a  day,  how  much  will  he  save 
in  365  days,  his  income  being  8400  ?  Ans.  8162.75 

12.  Bought  21  barrels  of  apples,  of  3  bushels  each, 
at  35  cts.  a  bushel :  what  did  they  cost?      Ans.  822.05 

13.  What  cost  four  pieces  of  calico,  each  containing  19 
yards,  at  23  cts.  a  yard?  Ans.  817.48 

14.  If  25  men  perform  a  piece  of  work,  for  82000,  and 
spend,  while  doing  it,  8163.75,  what  will  be  each  man's 
share  of  the  profits  ?  ^?is.  873.45 

15.  If  16  men  receive  8516  for  43  days'  work,  how 
much  does  each  man  earn  a  day  ?  Ans.  75  cts. 

16.  C  earned  890  in  40  days,  working  10  hours  a  day : 
how  much  did  he  earn  un  hour?  Ans.  22  cts.  5  m. 

3d  lik.  G 


82 


RAY'S    PRACTICAL    ARITHMETIC. 


17.  A  merchant,  failing,  has  goods  worth  SI 000,  and 
$500  in  cash,  to  be  equally  divided  among  22  creditors  : 
how  much  will  each  receive?  Aus.  S68.18-[- 

MEECHANTS'    BILLS. 

A  Bill  or  Account,  is  a  written  statement  of  articles 
bought  or  sold,  their  prices,  and  entire  cost. 

18.  Bought  9  pounds  Coffee,     at  $0.16  per  lb.       $ 

4  pounds  Tea,  ..     1.25      do. 

45  pounds  Sugar,      ..       .09      do. 
17  pounds  Cheese,    ..       .13      do. 


What  is  the  amount  of  my  bill  ? 


Ans.  $12.70 


19.  Bought  8  yards  Silk,  at  $1.10  per  yd.      $ 

18  yards  Muslin,     ..        .25      do. 

25  yards  Linen,       ..        .15      do. 

12  yards  Calico,      ..        .35      do. 

6  yards  Gingham,..        .65      do. 


What  is  the  whole  amount?  Ans.  $25.15 

20.  ancinnati,  Feb.  20th,  1860. 

Mr.  William  Ray,         t>       iv    ^Air  t?  a  e  r« 

'         Bought  of  W.  B,  omith  &  Co. 

5  Eclectic  Third  Readers,  at  $  0.35  each.         $ 

12  dozen  Olney,       10.50  per  doz. 

6000  Quills, 1.60  per  M. 

5  Quires  Paper, .25  per  quire. 

3  Copies  of  Hutton,   .     .      ..      4.50  each. 


21. 


liec^d  Payment, 


Mr.  John  Jones, 


27  Spelling  Books,    .     .     . 
25  Eclectic  Readers,      .     , 

8  Ainsworth's  Dictionaries, 
27  Greek  Readers,    .     .     . 
18  Bibles, 


W.  B.  Smith  &  Co.       ^152.10 

Boston,  July  5th,  1860. 

Bought  of  C.   Bradford, 
at  $0.19  each. 


.27 
4.50 
2.25 
1.50 


do. 
do. 
do. 
do. 
do. 


75  Testaments, 31 

Rec'd  Payment,  ^^  Bradfoiu). 


$i 


$158.88 
U^^  For  Fractional  Exercises  in  U.  S.  Money,  see  page  1G9. 


EEDUCTION  OF  COMPOUND  NUMBERS. 

Art.  77.  Reduction  is  the  process  of  changing  the 
denomination  of  a  number,  without  altering  its  value. 

Ex. — Since  3  feet  make  1  yard,  yards  may  be  changed  to  feet 
by  multiplying  by  3 ;     and,  feet  to  yards,  by  dividing  by  3  : 
5  yards  =  5X3  =  15  feet:     and  15  feet=  15-t- 3^5  yards. 

Hence,  as  shown  in  United  States  money. 

Reduction  Descending  consists  in  changing  a  number  from 
a  higher  to  a  lower  denomination :  Reduction  Ascending, 
in  changing  a  number  from  a  lower  to  a  higher  denomination. 

Rem. — The  Tables  teach  the  names  of  the  different  units,  and  the  num- 
ber of  units  of  one  order  or  denomination  which  make  a  unit  of  the  next 
higher  order :  they  are  analogous  to  the  Table  of  Orders  in  Simple  Numbers. 

Art.  78.  DRY    MEASURE 
Is  used  in  measuring  grain,  vegetables,  fruit,  coal,  &c. 

TABLE. 

2  pints    (pt.)        make  1  quart,         marked  qt 

8  quarts 1   peck,       pk. 

4  pecks     1  bushel, bu. 

The  standard  unit  of  dry  measure  is  the  Bushel ;  a  circular  measure 
of  ISj  inches  diameter,  8  inches  deep,  and  contains  2150S  cubic  inches. 

^^  Those  who  resort  to  Tables  of  Unit  Values,  to  avoid  each 
successive  step  of  an  operation,  lose  a  valuable  exercise. 

Notes. — 1.  There  are  two  other  denominations  of  Dry  Measure, 
the  quarter  and  chaldron.  The  quarter  contains  8  bushels,  of  70 
pounds  each,  used  in  England  in  selling  wheat. 

The  chaldron,  in  England,  and  in  some  of  the  U.  S.,  contains  36 
bu. ;  in  other  States  32  bu.,  and  is  used  for  measuring  coal, 

2.  When  Grain  and  Seeds  are  bought  and  sold  by  weight,  60  pounds 
of  Wheat,  60  of  Clover  seed,  56  of  Rye,  Corn,  or  Flax  seed,  32  of 
Oats,    42  of  Timothy  seed,    and  48  of  Barley,  make  1  bushel. 

For  information  in  detail  in  regard  to  foreign  and  domestic  weights  and 
measures,  see  "  Raxfs  Higher  Arithmetic." 

Review. — 77.  What  is  Reduction  ?  Give  an  example.  In  what  docs 
reduction  descending  consist  ?  In  what  does  reduction  ascending  consist  ? 
Rem.  What  do  the  Tables  t^ach  ? 

83 


84  RAY'S   PRACTICAL    ARITHMETIC. 

To  Teachers. — Numerous  questions  should  be  asked  on  each 
Table,  similar  to  the  following : 

1.  IIow  many  pints  in  1  quart?  in  2?  in  3?  in  4? 
in  5?  in  6?  in  7?  in  9?  in  10? 

2.  How  many  quarts  in  1  peck  ?  in  2  ?  in  3  ?  in  4  ? 
in  5?  in  6?  in  7?  in  8?  in  9?  in  10? 

3.  How  many  pecks  in  1  bushel?  in  2  ?  in  3?  in  4? 
in  5?  in  6?  in  7?  in  8?  in  9?  in  10? 

4.  How  many  quarts  in  2  pints?  in  5?  in  G?  in  8? 
in  9?  in  11?  in  12?  in  13?  in  14? 

5.  How  many  pecks  in  8  quarts?  in  16?  in  24?  in  35? 
in  40?  in  49?  in  56?  in  65? 

6.  How  many  bushels  in  4  pecks  ?  in  12  ?  in  20  ?  in  11  ? 
in  15?  in  27?  in  32?  in  39? 

Art.  79.  The  preceding  Examples  show,  that 

To  reduce  quarts  to  pints,  multiply  the  number  of  quarts 
by  the  number  of  pints  in  a  quart. 

To  reduce  pecks  to  quarts,  multiply  the  number  of 
pecks  by  the  number  of  quarts  in  a  peck. 

To  reduce  bushels  to  pecks,  multiply  the  number  of 
bushels  by  the  number  of  pecks  in  a  bushel.     Hence, 

Reduction  Descending  is  performed  by  MultipUcation :  the 
multiplier  being  that  number  of  the  lower  order  or  denomina- 
tioUy  which  makes  a  unit  of  the  next  higher. 

Art.  80.  To  reduce  pints  to  quarts,  divide  the  pints  by 
the  number  of  pints  in  a  quart. 

To  reduce  quarts  to  pecks,  divide  the  quarts  by  the 
number  of  quarts  in  a  peck. 

To  reduce  pecks  to  bushels,  divide  the  pecks  by  the 
number  of  pecks  in  a  bushel.     Hence, 

Reduction  Ascending  is  performed  by  Division  :  the  divisor 
being  that  number  of  the  loioer  order  or  denomination,  which 
makes  a  unit  of  the  next  higher. 

Review, — 78.  For  what  is  Dry  Measure  used?  Repeat  the  Table. 
What  is  the  ataiidard  unit  of  Dry  Measure  ? 


REDUCTION    OF    COMPOUND    NUMBERS. 


85 


7.  Reduce  3  bushels  to  pints. 

Solution. — To  reduce  bushels  to  pecks,  mul- 
tiply by  4,  because  there  are  4  pecks  in  a  bushel. 
To  reduce  pecks  to  quarts,  multiply  by  8,  be- 
cause there  are  8  quarts  in  a  peck,  or  8  times  as 
many  quarts  as  pecks.  To  reduce  quarts  to 
pints,  multiply  by  2,  because  there  are  2  pints 
in  a  quart. 


OPERATION. 

3  Bushels. 
4 


1 2  =  pk. 

8 


96=qt. 

2 


192  =  pt. 


8.  Reduce  192  pints  to  bushels. 

Solution. — To  reduce  pints  to  quarts,  di- 
vide by  2,  because  there  are  2  pints  in  a 
quart  To  reduce  quarts  to  pecks,  divide 
by  8,  because  there  are  8  quarts  in  a  peck. 
To  reduce  pecks  to  bushels,  divide  by  4,  be- 
cause there  are  4  pecks  in  a  bushel. 


OPERATION. 

2)192  Pints. 
8)96  =  qt. 
4)12  =  pk. 


The    two    preceding    examples    show    that 
Descending  and  Ascending  prove  each  other. 


3  =  bu. 
Redaction 


9.  Reduce  7  bushels,  3  pecks,  4  quarts,  1  pint,  to  pints. 


Solution. —  In  solving 
this  example,  multiply  the 
bushels  by  4,  which  make  28 
pecks ;  to  these  add  the  3 
pecks.  Then,  multiply  the 
pecks  (31)  by  8,  and  add 
the  4  quarts;  multiply  the 
quarts  (252)  by  2,  and  add 
the  1  pint. 


operation. 
Bu.  pk.  qt.  pt. 
7     3    4    1 
4 


31=  pk.  in  7  bu.  3  pk. 
8 


2  5  2  =  qt.  in  7  bu.  3  pk.  4  qt. 
2 


5  05  =  pints  in  the  whole. 


Review. — 78.  Note  1.  What  other  denominations  of  Dry  Measure  ? 
What  does  the  quarter  contain  ?  For  what  is  it  used  ?  What  does  the 
chaldron  contain  ?  For  what  is  it  used  ?  Note  2.  How  many  pounds  in 
a  bushel  of  wheat  ?     How  many  in  other  grains  ? 

79.  How  are  quarts  reduced  to  pints  ?  Pecks  to  quarts  ?  Bushels  to 
pecks  ?    How  is  reduction  descending  performed  ?     What  the  multiplier  ? 

80.  How  are  pints  reduced  to  quarts  ?  Quarts  to  pecks  ?  Pecks  to  bushels  ? 
How  is  reduction  ascending  performed  ?     What  is  the  divisor  ? 


86  RAY'S   PRACTICAL   ARITHMETIC. 

10.  Reduce  505  pints  to  bushels. 
Solution.— To   reduce  pints  operation. 

io  quarts,  divide  by  2,  and  there  Pt-  in  a  qt,  2)505 

is  1  left;  as  the  dividend  is  pints,  ^^    .^  ^  ^^  g  >^252qt.   1  pt. 

this  remainder  is  1  pint.  

To  reduce  quarts  to  pecks,  di-  Pk.  in  a  bu.     4)31  pk.  4  qt. 
vide  by  8,  and  4  quarts  are  left.  ~         0^.1, 

To  reduce  pecks  to  bushels,  di- 
vide by  4,  and  3  pecks  are  left.  ^^S-  ^  bu.  3  pk.  4  qt.    1  pt. 

The  remainder  is  always  of  the  same  denomination  as  the 
dividend,  Art.  38.     Hence, 

Art.  81.  GENERAL    RULES. 
To  Reduce  from  a  Higher  to  a  Lower  Order, 

Kule. — Multiply  the  highest  denomination  given,  by  that  num- 
ber of  the  next  lower ^  which  makes  a  unit  of  the  higher ;  add  to 
the  product  the  number,  if  any,  of  the  lower  denomination. 

Proceed  in  like  mariner  with  the  result  thus  obtained,  till  the 
whole  is  reduced  to  the  required  denomination. 

To  Reduce  from  a  Lower  to  a  Higher  Order, 

Rule. — Divide  the  given  quantity  by  that  ruimber  of  its  oicn 
denomination  which  makes  a  unit  of  the  next  higher. 

Proceed  in  like  manner  with  the  quotient  thus  obtained,  till 
the  whole  is  reduced  to  the  required  denomination. 

The  last  quotient,  with  the  several  remainders,  if  any,  annexed, 
will  be  the  answer. 

Proof. — Reverse  the  operation :  that  is,  reduce  the  answer 
back  to  the  denomination  from  which  it  was  derived.  If  this 
result  is  the  same  as  the  quantity  given,  the  work  is  correct 

11.  Reduce  2  bu.  to  pints Ans.  128  pt. 


12.  12  pk.  to  pints. 

13.  8  bu.  to  quarts. 

14.  1  bu.  1  pk.  to  pints. 

15.  2  bu.  2  qt.   to  pints. 


Ans.  192  pt. 
Ans.  256  qt. 
Ans.  80  pt. 
Ans.  132  pt. 


Review. — 81.  "What  is  the  general  rule  for  reducing  from  a  higher  to  a 
lower  order  ?    From  a  lower  to  a  higher  ?     What  the  method  of  proof? 


4 


REDUCTION   OF    COMPOUND    NUMBERS.  87 

16.  Reduce  4  bu.  2  pk.  1  qt.  to  pints.     Ans.  290  pt. 

17.  7  bu.  3  pk.  7  qt.  1  pt.  to  pt.       Ans.  511  pt. 

18.  3  bu.  1  pt.  to  pints.       .     .     .     Ans.  193  pt. 

19.  384  pt.  to  bushels.  .     .     .     Ans.      6  bu. 

20.  47  pt.  to  pecks.  Ans.  2  pk.  7  qt.  1  pt. 

21.  95  pt.  to  bu.        Ans.  1  bu.   1  pk.  7  qt.  1  pt. 

22.  508  pt.  to  bu.  Ans.  7  bu.  3  pk.  6  qt. 

•Art.  82.   TROY    OR    MINT   WEIGHT 
Is  used  in  weighing  gold,  silver,  jewels,  liquors,  &c. 

TABLE. 

24  grains  (gr.)       make  I  pennyweight,  marked  pwt. 

20  pennyweights     ...    1  ounce,     oz. 

12  ounces 1  pound,    lb. 

Note. — The  standard  wiit  of  weight  in  the  United  States,  is  the  IVoy 
prmnd,  containing  5760  grains. 

For  interesting  historical  and  other  information  with  respect  to  coins, 
see  "  Bay's  Higher  Arithmetic." 

Teachers  should  ask  questions  on  each  Table,  as  on  Dry  Measure. 

1.  Reduce  13  lb.  11  oz.  16  pwt.  14  gr.  to  grains. 

Suggestion. — When  the  de-  167  oz.  {Brought  up.) 

nominations  to  be   added  are  2  0 

small,  add  while  multiplying;  " 

when  large,  beginners  should  oo4:U  pwt. 

add  after  multiplying.  1^  pwt.  to  be  added. 


OPERATION. 

lb.  OZ.  pwt.  gr. 
13  11  16  14 
12 


3356  pwt. 
24 


13424 
6712 


156  oz.  80544  gr. 

11  oz.  to  be  added.  1 4  g^.  to  be  added. 

1  6  7  oz.  ( Carried  up.)  80558  gr. 


Review.— 82.  For  what  is   Troy  "Weight  used?    Repeat  the  Table. 
Note.  What  is  the  standard  unit  of  weight  in  the  United  States  ? 


88 


RAY'S   PRACTICAL  ARITHMETIC. 


2. 
3. 
4. 
5. 

6. 

7. 

8. 

9. 
10. 
11. 


Reduce  4  lb.  to  grains. 


rrx 


.     .     .     Ans.  23040 
5  lb.  4  oz.  to  ounces.        .     .     .  Ans.  64  oz. 

9  lb.  3  oz.  5  pwt.  to  pwt.  .  .  Ans.  2225  pwt. 
141b.  11  oz.  19  pwt.  23  gr.  to  gr.  Ans.  86399  gr. 
8  lb.  9  oz.  13  pwt.  17  gr.  to  gr.  Ans.  50729  gr. 
171  gr.  to  pennyweights.  Ans.  7  pwt.  3  gr. 

505  gr.  to  ounces.  Ans.  1  oz.  1  pwt.  1  gr. 

12530  gr.  to  pounds.  Ans.  2  lb.  2  oz.  2  pwt.  2  gr. 
805  pwt.    to  pounds.  Ans.  31b.  4oz.«5pwt. 

25591  gr.  to  pounds.  Ans.  4  lb.  5  oz.  6  pwt.  7gr. 


Art.  83.   APOTHECARIES    WEIGHT 
Is  used  by  Apothecaries  in  compounding  medicines. 


TABLE. 

20  grains  (gr.)     make  1   scruple, 

3  scruples 1  dram,    . 

8  drams       1  ounce, 

12  ounces      1  pound, 


marked  3. 


5- 

tb. 


2. 
3. 
4. 

5. 
6. 
7. 
8. 
9. 
10. 


.  Ans.  17280  gr. 

.  Ans.  23342  gr. 

.  Ans.     2018  9. 

.  Ans.  41300  gr. 

.  Ans.    51b  7  5- 

^«s.  4tb  5575. 

975  9  to  pounds i4ws.  31b  45  5  5. 

6321  gr.  to  pounds.  Ans.  1  lb  1  5  1  5  1  9  1  gr. 
30941  gr.  to  pounds.  Ans.  51b  4  5  3  3  29  1  gr. 
29239  gr.  to  pounds.     .     .     Ans.  5  lb  7  5  19  gr. 


4  lb  55  2  gr.  to  grains. 
7  tb  2  9  to  scruples. 
7  lb  2  3  1  9  to  grains. 
67  5  to  pounds. 
431  5  to  pounds.     .     . 


Review. — 83.  For  what  is  Apothecaries  "Weight  used?  Repeat  the 
Table,  What  is  said  of  the  grain,  ounce,  and  pound,  of  Apothecaries  and 
Troy  weight?     What  is  differently  divided  ? 


The  grain,  ounce,  and  pound,  in  Apothecaries  and  Troy  weight,  are  the 
same  :  but  the  ounce  is  differently  divided.  j 

^^  Questions  should  be  asked  on  the  Table,  as  before. 

1.  Reduce  31b  to  o:rains. 


REDUCTION    OF    COMPOUND   NUMBERS.  89 

Art.  84.   AVOIRDUPOIS    WEIGHT 

Is   used    in   weighing   heavy   articles  ;     as,    groceries, 
coarse  metals   and  medicines  at  wholesale. 

TABLE. 

16  drams  (dr.)  .    .    .  make  1  ounce,.    .  marked  oz. 

16  ounces 1  pound, lb. 

25  pounds     . 1  quarter,      .    .    .    .   qr. 

4  quarters  or  100  lb.    .    .    1  hundred  weight,   .   cwt 
20  hundred  weight      ...    1  tun, T. 

Notes. — 1.  The  standard  Avoirdupois  pound  of  the  United  States  is 
determined  from  the  Troy  pound,  and  contains  7000  grains  Troy. 

2.  Formerly,  28  pounds  were  allowed  for  a  quarter,  112  pounds  for  a 
hundred  weight,  2240  jwunds  for  a  tun  ;  these,  called  the  long  hundred 
and  long  tun,  are  chiefly  used  at  the  Custom  House. 

1.  Reduce  2  cwt.  to  pounds.  .  .  .  Ans.    200  lb. 

2.  3  cwt.  3  qr.  to  pounds.   .  .  .  Ans.    375  lb. 

3.  1 T.  2  cwt.  to  pounds.    .  .  .  Ans.  2200  lb. 

4.  3  T.  3  qr.  to  pounds.    .  .  .  Ans.  6075  lb. 

5.  4  cwt.  1  qr.  191b.  to  pounds.   .  Ans.    444  1b. 

6.  5T.  3qr.  15  lb.  to  pounds.  .  .  Ans.  10090  lb. 

7.  2  cwt.  3  qr.  2  lb.  12  oz.  to  ounces.  Ans.    4444  oz. 

8.  2  cwt.  17  lb.  3  dr.  to  drams.  .  .  Ans.  55555  dr. 

9.  IT.  6 cwt.  41b.  2oz.  10 dr.  to  dr.  Ans.  666666  dr. 

10.  48031b.  to  cwt Ans.  48  cwt.  31b. 

11.  22400  lb.  to  tuns ^ns.  11  T.  4  cwt. 

12.  2048000  dr.  to  tuns.     .     .     .  Ans.  4  T. 

13.  64546  dr.  to  cwt.     Ans.  2  cwt.  2  qr.  2  lb.  2  oz.  2  dr. 

14.  97203  oz.  to  tuns.  Aris.  3T.  3qr.  3oz. 

15.  544272 dr.  to  T.     Ans.  IT.  Icwt.  Iqr.  1  lb.  loz. 

16.  In  52  parcels  of  sugar,  each  containing  18  lb.,  how 
many  hundred  weight?  Ans.  9  cwt.  1  qr.  11  lb. 

Review. — 84,  For  what  is  Avoirdupois  Weight  used  ?  Repeat  the 
Table.  Note  1.  From  what  is  the  standard  Avoirdupois  pound  deter- 
mined ?    2.  What  is  the  lon-g  hundred  and  long  tun  ?    Where  used  ? 


90 


RAY'S   PRACTICAL    ARITHMETIC. 


Art.  85.  LONG    OK    LINEAR    MEASURE: 

Used  in  measuring  lengths,  breadths,  thickness,  &c. 

TABLE. 

12    inches  (in.).    .   make  I  foot,      .    .   marked  ft. 

3    feet      1  yard, yd. 

5^  yards  or  1 6^  feet  .    .    1  rod,  perch,  or  pole,  rd. 

40    rods  or  220  yards   .    1  furlong, fur. 

8    furlongs  or        )             ,      •,  . 

1760    yd.  =  5280ft.  I    *   *   "^  ^^^^' ^^ 


ALSO,   60  geographic,  or  69^  statute  miles,  make  1  degree. 

360  degrees  make  a  great  circle,  or  circumfereuce  of  the  earth. 

3  miles  make  1  league,  used  in  measuring  distances  at  sea. 

4  inches  make  1  hand,  used  in  measuring  the  hight  of  horses. 
6  feet  make  1  fathom,  used  in  measuring  the  depth  of  water. 

Note. — The  stafulard  unit  of  length  is  the  yai'd. 


1.  Reduce  2yd.  2ft.  Tin.  to  inches. 

2.  7  yd.  11  in.  to  inches. 

3.  12  mi.  to  rods.       .     . 

4.  7  mi.  6  fur.  to  rods. 

5.  9  mi.  31  rd.  to  rods. 

6.  133  in.  to  yards. 

7.  181  in.  to  yards. 

8.  2240  rd.  to  miles.       . 

9.  2200  rd.  to  miles.      . 


Ana.  103  in. 
.  .  Ans.  263  in. 
.  .  ^«s.  3840rd. 
.  .  A71S.  2480  rd. 
.  .  Ans.  2911  rd. 
Ans.  3  yd.  2  ft.  1  in. 
Ans.  5  yd.  1  in. 
,  .  .  Alls.  7  mi. 
Ans.  6  mi.  7  fur. 


For  examples  involving  Fractions,  see  page  170. 


Art.   86.   LAND    OR    SQUARE    MEASURE 

Is  used  in  measuring  land,  or  any  thing  in  which  both 
length  and  breadth  are  considered. 


Art.  87.  A  figure  having  four  equal  sides, 
and  four  right  angles  (corners),  is  a  square. 

Hence,  A  square  inch  is  a  square,  each  side  of 
which  is  a  linear  inch;  that  is,  1  inch  in  length. 


1  inch. 


ono 

square 

inch. 


Kkview, — 85.  For  what  is  Long  Measure  used?    Repeat  the  Table. 
Note.  What  is  the  standard  unit  of  length  ? 


REDUCTION   OF    COMPOUND  NUMBERS. 


91 


CO 


one 
eq.  ft. 

1 

A  square  foot  is  a  square,  each  side  of  which  is  a  linear  foot. 

3  feet.  A  square  yard  is  a  square,  each  side  of 

which  is  a  linear  yard  (3  feet). 

The  figure  shows  that  1  square  yard,  that 
is,  3  feet  square,  contains  9  square  feet. 

The  number  of  small  squares  in  any 
large  square,  is  equal  to  the  number  of  units 
in  one  side  of  the  large  square  multiplied 
by  itself     Thus, 

In  a  square  figure,  each  side  of  which  is  8  inches,  there 
are  64  square  inches :  in  1  foot  square,  or  12  inches  on  each 
side,  there  are  12X  12  =  144  square  in. 

Art.  88.  By  3  inches  square,  we  mean 
a  square  figure,  each  side  3  inches  :  but, 

3  square  inches  are  3  small  squares,  each 
an  inch  long,  and  an  inch  wide ;  and, 

3  inches  square  contain  9  square  inches. 

The  difi'erence   between  4  in.  square,  and  4       3  square  in. 
square  in.,  is  12  square  inches;  between  6  miles 
square,  and  6  square  miles,  is  30  sq.  miles. 

TABLE. 
144    square  inches    make  1  square  foot,    marked  sq.  ft. 


3  in. squ 

are. 

9  square  feet  . 
SQi  square  yards 
40    perches      .    . 

4  roods  .  .  . 
640    acres     .   .    . 


ALSO,  Ij^Q  inches 


4  rods  or  66  feet 
80  chains     .    .    . 

1  square  chain 
10  square  chains 


1  square  yard,     .    .    .  sq.  yd. 

1  sq.  rod  or  perch,  .    .  P. 

1  rood,      R. 

1  acre,      A 

1  square  mile,      .    .    .  sq.  mi. 

1  link, L 

1  chain, ch. 


1  mile,       mi. 

16  perches, P. 

1  acre,       A. 

Note. — Land  is  measured  with  a  surveyor's  or  Gunter's  chain  ;  it  is 
4  rods  or  66  feet  in  length,  and  is  divided  into  100  links. 


Eeview, — 86.  For  what  is  Land  or  Square  Measure  used?  87.  What 
is  a  square  ?  What  is  a  square  inch  ?  A  square  foot  ?  A  square  yard  ? 
To  what  is  the  number  of  small  squares  in  a  large  square  equal  ? 


92 


RAY'S   PRACTICAL    ARITHMETIC. 


1. 
2. 
3. 

4. 
5. 

6. 
7. 
8. 
9. 


Reduce  8sq.  yd.  to  square  inches.  Ans.  10368 sq.  in. 

4  A.  to  perches Ans.  640  P. 

Isq.  mi.  to  perches Ans.  102400  P. 


2  sq.  yd.  3  sq.  ft.  to  sq.  in. 
5  A.  2  R.  20  P.  to  perches. 
960  P.  to  acres.       .     .     . 
3888  sq.  in.  to  square  yards. 
243  P.  to  acres.        .     .     . 
603  P.  to  acres.       .     .     . 


Ans.  3024  sq.  in. 

.     .     Ans.  900  P. 

.     .     .    Ans.  6  A. 

.  .  Ans.  3  sq.  yd. 
Ans.  1  A.  2  R.  3  P. 
Ans.  3  A.  3  R.  3  P. 


10.     4176  sq.  in.  to  sq.  yd.        .    ^/is.  3  sq.  yd.  2  sq.ft. 

Art.  89.  A  Rectangle  is  a  figure  having  four  sides  and 
four  right  angles.     See  the  figure  below. 

The  Area  or  Superficial  content  of  a  figure,  is  the  number  of 
times  it  contains  its  U7iit  of  measure. 

The  unit  of  measure  for  surfaces,  is  a  square  whose  side  is  a 
linear  unit,  as  a  square  inch,  a  square  foot,  &c. 

1.  How  many  square  inches  in  a  board  4  inches  long 
and  3  inches  wide  ? 


Sol. — Dividing  each  of  the  longer  sides  into  4 
equal  parts,  the  shorter  sides  into  3  equal  parts, 
and  joining  the  opposite  divisions  by  straight 
lines,  the  surface  is  divided  into  squares. 

In  each  of  the  longer  rows  there  are  4  squares,  that  is,  as  many 
as  there  are  inches  in  the  longer  side;  and  there  are  as  many  such 
rows  as  there  are  inches  in  the  shorter  side.     Hence, 

The  whole  number  of  squares  in  the  board  is  equal  to  the  pro- 
duct obtained  by  multiplying  together  the  numbers  representing 
the  length  and  breadth;  that  is,  4  X  3  =  12.     Hence  the 

Rule  for  Finding  the  Area  of  a  Rectangle. — Multiply  the 
lengthhy  the  breadth:  the  product  will  he  the  superficial  contents. 

Ekview. — 88.  What  is  meant  by  8  inches  square  ?  By  3  sq.  inches  ? 
What  is  the  diflference  between  4  inches  square,  and  4  square  inches? 

89.  What  is  a  Rectangle  ?  What  is  the  Area  of  a  figure  ?  What  is  the 
unit  of  mcn,«nro  for  surfaces  ? 


REDUCTION   OF   COMPOUND   NUMBERS.  93 

Note. — Both  the  length  and  breadth,  if  not  in  units  of  the  game 
denomination,  must  be  made  so,  before  multiplying. 

2.  In  a  floor  16  feet  long  and  12  feet  wide,  how  many 
square  feet  ?  J  us.  192  sq.  ft. 

3.  How  many  square  yards  of  carpeting  will  cover  a 
room  5  yds.  long  and  4  yds.  wide  ?  Ans.  20  sq.  yd. 

4.  How  many  square  yards  of  carpeting  will  cover  two 
rooms,  one  18  feet  long  and  12  feet  wide,  the  other  21  feet 
long  and  15  feet  wide  /  Ans.  59  sq.  yd. 

5.  How  many  square  yards  in  a  ceiling  18  feet  long 
and  14  feet  wide  ?  Ans.  28  sq.  yd. 

6.  In  a  field  35  rods  long  and  32  rods  wide,  how  many 
acres?  Ans.  7  A. 

Art.  90.  The  Area  of  a  Rectangle  being  equal  to  the 
product  of  the  length  by  the  breadth,  and  as  the  product 
of  two  numbers,  divided  by  either  of  them,  gives  the 
other  (Art.  S7)  ;  therefore. 

Rule. — If  the  area  of  a  rectangle  he  divided  hy  either  side, 
the  quotient  icill  he  the  other  side. 

Illustration. — In  Example  1,  Art.  89,  if  the  area,  12,  be 
divided  by  4,  the  quotient,  3,  is  the  width ;  or,  divide  12  by  3, 
the  quotient,  4,  is  the  length. 

Note. — Dividing  the  area  of  a  rectangle  by  one  of  its  sides,  is  really 
dividing  the  number  of  squares  in  the  rectangle  by  the  number  of  squares 
on  one  of  its  sides.     Thus, 

In  dividing  12  by  4,  the  latter  is  not  4  linear  inches,  but  the  number  of 
inches  in  a  rectangle  4  in.  long  and  1  in.  wide.     See  figure,  Art.  89. 

1.  A  floor  containing  132  square  feet,  is  11  feet  wide  : 
what  is  its  length  ?  Ans.  12  ft. 

2.  A  floor  is  18  feet  long,  and  contains  30  square  yards  : 
what  is  its  width?  Ans.  15  ft. 

3.  A  field  containing  9  acres,  is  45  rods  in  length : 
what  is  its  width  ?  Ans.  32  rd. 

4.  A  field  35  rods  wide,  contains  21  acres  :  what  is  its 
length?  Ans.  96 rd. 

Eeview. — 89.  "What  is  the  Eule  for  finding  the  Area  of  a  Eectangle  ? 
90.  The  area  and  one  side  being  given,  huw  m:iy  the  other  side  be  found  ? 


94 


RAY'S  PRACTICAL  ARITHMETIC. 


Art.  91.  SOLID   OR  cubic  measure: 

Used  in  measuring  things  having  length,  breadth,  and 
thickness;  as,  timber,  stone,  earth. 

Art.  92.  A  Cube  is  a  Solid,  hav- 
ing 6  equal  faces,  which  are  squares. 

If  each  side  of  a  cube  is  1  inch  long, 
it  is  called  a  cubic  inch ;  if  each  side 
is  3  feet  (1  yard)  long,  as  in  the  figure, 
it  is  a  cubic  or  solid  yard. 

The  base  of  a  cube,  being  1  square  3  feet  long,    "b 

yard,  contains  3X3  =  9  square  feet;  and  1  foot  high  on  this  base, 
contains  9  solid  feet;  2  feet  high  contain 9  X  2  =  18 solid  feet;  3  feet 
high  contain  9X3=27  solid  feet.  Also,  it  maybe  shown  tlint  1  solid 
or  cubic  foot  contains  12X12X12  =  1728  solid  or  cubic  inches. 

Hence,  the  number  of  small  cubes  in  any  large  cube,  is  equal  to 
the  length,  breadth,  and  thickness,  multiplied  together. 

Art.  93.  Any  solid,  whose  corners  resemble  a  cube,  is 
a  rectangular  solid  :  boxes  and  cellars  are  of  this  form. 

The  solid  contents  of  a  rectangular  solid  are  found,  as  in  the 
cube,  by  multiplying  together  the  length,  breadth,  and  thickness. 

TABLE. 

1728  cubic  inches  (cu.  in.)     make  1  cubic  foot,  marked  cu.  ft. 

27  cubic  feet 1  cubic  3'ard,  .    .    .   cu.yd. 

40  feet  of  round  or         )  •.  .  . 

50  feet  of  hewn  timber  )    '    *    *  ' 

128  cubic  fect-=8x4X4  =  8  ft.  )  ^       j  ^ 

long,4ft.  wide,  and  4it  high,  j  '    ' 

Note. — A  cord  foot,  is  1  foot  in  length  of  the  pile  which  makes  a  cord. 
It  Is  4  feet  wide,  4  feet  high,  and  I  foot  long  ;  hence,  it  contains  16  cubio 
feet,  and  8  cord  feet  make  1  cord. 


1.  Reduce  2 cu.yd.  to  cubic  inches. 

2.  28  cords  of  wood  to  cu.ft. 


Ans.  9331 2 cu.  in. 
Ans.  3584  cu.  ft. 


Review. — 91.  For  whnt  is  Cubic  Measure  used?  92.  What  is  a  cube? 
A  cubic  inch  ?  Cubic  yard  ?  How  m:»ny  sq.  ft.  in  one  side  of  a  cubio 
y:ird  ?  How  many  cubic  feet  in  a  solid  3  ft.  long,  3  ft.  wide,  and  1  ft 
thick  ?     If  2  ft.  thick  ?     Ii"  3  ft.  thick  ? 


REDUCTION  OF   COMPOUND   NUMBERS.  95 

3.  Reduce  34  cords  of  wood  to  C.  ft.      Ans.  272  C.  ft. 

4.  1  cord  of  wood  to  cu,  in.  Ans.  22118-i  cu.  in. 

5.  63936  cu.  in.  to  cu.  yd.        Ans.  1  cu.  yd.  10  cu.  ft. 

6.  492480  cu.  in.  to  tn.  round  timber.  Ans.ltn.  5cu.ft. 

7.  How  many  cubic  feet  in  a  rectangular  solid,  8  ft. 
long,  5  ft.  wide,  4  ft.  thick  ?    See  Art.  93.     A]is.  160  cu.  ft. 

8.  How  many  cubic   yards  of  excavation  in  a  cellar  8 
yd.  long,  5  yd.  wide,  2  yd.  deep?  Ans.  80  cu.  yd. 

9.  How  many  cubic  yards  in  a  cellar,  18  feet  long,  15 
feet  wide,  7  feet  deep  ?  Ans.  70  cu.  yd. 

10.  In  a  pile  of  wood  40  feet  long,  12  feet  wide,  and  8 
feet  bigb,  how  many  cords  ?  Ans.  30  C. 

Art.  94.  CLOTH   MEASURE 
Is  used  in  measuring  cloth,  muslin,  ribbons,  tape,  &c. 

p  TABLE. 

*         2\  inches  (in.)       make  1  nail,      .    .    .   marked  na. 
4    nails  (or  9  in.)    .    .    1  quarter  of  a  yard,        qr. 

4  quarters  (3(*  in.)    .1  yard,       yd. 

5  quarters 1  Ell  Flemish,  .    .    .    .   E.  Fl. 

5  quarters 1  Ell  English,  .    .    .    .   E.  En. 

6  quarters 1  Ell  French,   .    .    .    .  E.  Fr. 

This  is  a  species  of  long  measure,  the  yard  being  36  inches  in  both. 

1.  Reduce  19  yd.  to  nails Ans.  304  na. 

2.  14E.ri.  to  nails Ans.  16S  na,. 

3.  5  yd.  2  qr.  3  na.  to  nails.      .     .     .     Ans.    91  na. 

4.  13  E.  En.  1  qr.  to  nails.       .     .     .     Ans.  264  na. 

5.  23  E.  Fr.  3  qr.  2  na.  to  nails.     .     .     Ans.  566  na. 

6.  159  na.  to  yards.    .     .     .     Ans.  9  yd.  3qr.  3  na. 

7.  287  na.  to  E.  Fr.    .     .     Ans.  11  E.  Fr.  5  qr.  3  na. 

8.  6  yd.  to  Ells  Flemish Ans.  8  E.  Fl. 

Reduce  the  yards  to  quarters  ;  then  the  quarters  to  Ells  Flem. 

Review. — 92.  To  what  is  the  number  of  small  cubes  in  any  large  cube 
equal?  93.  What  is  a  rectangular  solid?  How  is  the  solid  conienrs 
found  ?  Repeat  the  Table.  94.  For  what  is  Cloth  Measure  used?  Repeat 
the  Table.     Of  what  is  Cloth  Measure  a  species  ? 


96 


RAY'S  PRACTICAL   ARITHMETIC. 


9.  Reduce  9  yd.  3qr.  to  Ells  Flemish.    Ans.  13E.F1. 


10.  12  yd.  Iqr.  to  E.  Fl.     . 

11.  37  E.  Fl.  to  yards.     .     . 

12.  36  E.  En.  to  Ells  French. 

13.  22  E.  Eu.  4  qr.  to  Ells  Fr 

14.  47  E.  Eu.  to  yards. 


Aus.  16  E.  Fl.  1  qr. 
.  Ans.  27  yd.  3  qr. 
.  .  Ans.  30  E.  Fr. 
.  .  Ans.  19  KFr. 
,     Ans.  58  yd.  3  qr. 


Art.  95.  WINE    OR    LIQUID    MEASURE: 

For  measuring  all  liquids  except  beer,  milk,  and  ale. 

TABL  E. 

4  gills  (gi  )         make  1  pint,         marked  pt 

2  pints       1  quart,      .    .    .    .   qt 

4  quarts 1  gallon,    ....   gal. 

63  gallons 1  hogshead,  .    .    .   hlid. 

4  hogsheads      ...    1  tun, T. 

Also,  31i  gallons 1  barrel,    ....  LI. 

42 ~  gallons  .....  1  tierce,     .    .    .    .  tr. 

84    gallons i  puncheon,      .    .  pn. 

12G    gallons 1  pipe, p. 

The  Standard  Unit  of  Liquid  Measure  is  a  gallon  of  231  cubic  inches. 


1.  Reduce  17  gal.  to  pints. 

2.  13  gal.  to  gills.     .     . 
2  hhd.  to  pints.     .     . 


3. 

4. 

5. 

6. 

7. 

8. 

9. 
10. 
11. 
12. 


5  T.  to  gills. 


3  T.  3  hhd.  to  gallons.    . 
Ihhd.  GO  gal.  1  pt.  to  pints 
2  hhd.  17  gal.  3qt.  to  gills 
2T.  62  gal.  Ipt.  to  gills. 
96  iri.  to  callous. 


6048  gi.  to  hogsheads. 
32256  ffi.  to  tuns.     . 


4050  gi.  to  hogsheads.      Ans 


Ans.  136  pt. 
Ans.  416  gi. 
Ans.  1008  pt. 
Ans.  40320  gi. 
Ans.  945  gal. 
Ans.  985  pt. 
Ans.  4600  gi. 
Ans.  18116  gi. 
.    Ans.  3  gal, 
.    Ans.  3  hhd. 
.    Ans.  4:T. 
2  hhd.  2qt.  2gi. 


Reviicw. 95.  For  what  is  Liquid  Measure  used  ?     Repeat  the  Table. 

Note.  What  is  the  standard  unit  of  Liquid  Meaijure  ? 


REDUCTION    OF    COMPOUND   NUiMBERS.  97 

13.  Reduce 30339 gi.  to  T.    Am.ST.  3hhd.  3gal.  3gi. 
U.     10125gi.toT.    ^»s.  IT.  Ihhd.  Igal.  Iqt.  Ipt.  Igi. 

15.  3  puncheons  to  gills Ans.  8064  gi. 

16.  5  pipes  to  quarts Ans.  2520  qt. 

17.  5712  pt.  to  tierces Ans.  17  tr. 

Art.  96.   ALE    OR    BEER    MEASURE 
Is  generally  used  in  measuring  ale,  beer,  and  milk. 

TABLE. 

2  pints  (pt.)         make  1  quart,     marked  qt. 

4  quarts      1  gallon,  ....   gal. 

36  gallons 1  barrel,  .    .    .    .   bl. 

54  gallons 1  hogshead,     .    .   hhd. 

Note. — The  beer  gallon  contains  282  cubic  inches.     In  most  places 
milk  is  now  sold  by  Wine  Measure. 

1.  Reduce  4 hhd.  to  pints Ans.  1728  pt. 

2.  7  hhd.  3  qt.  to  pints ^tjs.  3030pt. 

3.  1000 pt.  to  barrels Ans.  3bl.  17 gal. 

4.  443 pt.  to  hhd.      .       Ans.  Ihhd.  Igal.  Iqt.  Ipt. 

Art.  97.  TIME    MEASURE. 

Time  is  a  measured  portion   of  duration.     The  parts 
into  which  time  is  divided  are  shown  in  this 

TABLE. 

GO  seconds  (sec.)        make  1  minute,     marked  min. 

60  minutes 1  hour, hr. 

24  hours 1  day,      da. 

365  days  6  hours,  {365ida.)  1  solar  year,  .  .  .  yr. 
100  years 1  century,  ....   cen. 

Also,    7  days 1  week, wk 

4  weeks     ....     ...  1  month,  (nearly)     mon. 

12  calendar  months  ...  1  year, yr. 

365  days 1  common  year,     .  yr. 

366  days 1  leap  year,    .    .    .  yr. 

^^^  See  Notes  on  page  98. 

Review.— 96.  For  what  is  Beer  Measure   used?     Repeat  the   Table. 
97.  What  is  Time  ?     Repeat  the  Table. 

3d  Bk.  7 


98 


RAY'S    PRACTICAL   ARITHMETIC. 


Note  1.  The  exact  length  of  the  solar  year  is  365  days,  5 
hours,  48  minutes,  48  seconds:  but,  is  usually  considered  to  be 
365  days,  6  hours.     Hence, 

One  year  being  regarded  as  365  days,  6  hours,  the  odd  6  hours 
of  each  year,  make,  in  4  years,  24  hours,  an  additional  day.  This 
gives  366  days  to  every  fourth  year,  called  leap  year;  and. 

All  leap  years  may  be  exactly  divided  by  4;  thus,  1848,  1852, 
1856,  are  leap  years;  while  1847,  1850,  1855,  are  not  so:  and. 

The  additional  day  in  leap  yeai's  is  added  to  February,  making 
this  month  29  days,  instead  of  28,  as  in  common  years. 

Note  2. — The  diflference  between  the  solar  year  and  365  days 
and  6  hours,  is  about  three-fourths  of  a  day  in  100  years,  or  3  days 
in  400  years.     Hence, 

For  greater  accuracy,  it  is  agreed  that  of  the  centennial  (hundredth) 
years,  only  those  which  are  exactly  divisible  by  400  shall  be  leap  years: 
thus,  1900  will  be  a  common  year,  and  2000  a  leap  year. 

Teachers  and  advanced  students  will  find  interesting  and  instructive 
historic  information  on  this  subject  in  "  Hay's  Higher  Arithmetic." 

Note  3.  The  year,  as  used  by  civilized  nations,  is  divided  into 
12  calendar  months,  and  numbered  in  their  order  as  follows: 

January,    lstmonth,31  days.    July,  7th  month,  31  days. 


February,  2d 

.      28    .. 

August,        8th 

.       31 

March,       3d 

31     ,. 

September,  9th 

.       30 

April,        4th      . 

.      30    .. 

October,     10th 

.       31 

May,          5th      . 

.       31     .. 

November,  11th 

.       30 

June,         Cth 

30    .. 

December,  12th 

.       31 

Thirty  days  hath  September,  April,  June,  and  November : 

All  the  rest  have  thirty-one,  save  February,  which  alone 

Hath  twenty-eight ;  and  one  day  more  wc  add  to  it,  one  year  in  four. 

1.  Reduce  2hr.  to  seconds.     .     .     .      Ans.  7200  sec. 

2.  7  da.  to  minutes Ans.  10080  min. 

3.  Ida.  3hr.  44 min.  3 sec.  to  sec.       Ans.  99843 sec. 

4.  9wk.  6  da.  10 hr.  40  min.  to  min.  Ans.  100000 min. 


Review. — 97.  Note  1.  What  is  the  exact  length  of  the  solar  year? 
What  are  leap  years?  February  has  how  many  dnys  in  leap  years? 
2.  Which  of  the  centennial  years  are  leap  years?  Why?  3.  How  is  the 
year  divided  ?     Give  the  number  of  days  in  each  mouth. 


REDUCTION   OF    COMPOUND  NUMBERS. 


99 


5.  Reduce  Imon.  3da.  4min,  to  min.     ^?is.  44644 min. 

6.  Reduce  lyr.  20  da.  IDhr.  15  min.  33 sec.  to  seconds, 
allowing  365  days  to  a  year.  Ans.  33333333  sec. 

7.  How  many  seconds  in  an  exact  solar  year  ? 

(See  Note  1,  page  98.)  Ans.  31556928 sec. 

8.  Reduce  10800  sec.  to  hours.       .     .     .      Ans.  31ir. 

9.  432000  sec.  to  days.  ....      Ans.  5  da. 

10.  7322  sec.  to  hours.      .     Ans.  2hr.  2  min.  2  sec. 

11.  4323 min.  to  days.       .     .     .     Ans.  3 da.  3 min. 

12.  20280  min.  to  weeks.       .     .     ^?is.  2wk.  2hr. 

13.  41761  min.  to  mon.        Ans.  Imon.  Ida.  Imin. 


Art.  98.   CIRCULAR    MEASURE 

Is  used  in  estimating  latitude  and  longitude,  and  in 
measuring  the  motions  of  the  heavenly  bodies. 

A  circle  is  divided  into  360  equal  parts,  called  degrees;  each 
degree  into  60  parts,  called  minutes;  each  minute  into  60  parts, 
called  seconds. 

TABLE. 


60  seconds  (/^)        make  1  minute, 
00  minutes       1  degree, 


marked  ''. 

o 


1. 

2. 
3. 
4. 
5. 

6. 

7. 


30  degrees 

12  signs  or  360° 

Reduce  5°  3'  to  minutes. 
8°  41'  45''  to  seconds 
3^  25'  to  minutes 
1^  to  seconds. 
244"  to  minutes. 
915'  to  deo-rees. 


1861'  to  signs. 


1  sign, 

1  circle, c. 

Ans.  303'. 

.    ^ns.  31305". 

.      Ans.  5425'. 

Ans.  1296000". 

Ans.  4'  4". 

Ans.  15°  15'. 

Ans.  13  1°  1'. 


Art.  99.  miscellaneous  table. 

24  sheets  of  paper make  1  quire. 

20  quires 1  ream. 

2  reams 1  bundle. 


Review. — 98.  For   what   is    Circular   Measure   used?     How   is   every 
Circle  divided  ?     Repeat  the  Table. 


100  RAY'S   PRACTICAL   ARITHMETIC. 

OP     BOOKS. 

A  sheet  folded  in    2  leaves  is  called  a  folio. 

A  sheet  folded  in    4 a  quarto,  or  4to. 

A  sheet  folded  in    8 an  octavo,  or  8vo. 

A  sheet  folded  in  12 a  duodecimo,  or  12ino. 

A  sheet  folded  in  18 an  ISmo. 

OP    THINGS. 

12  things make  1  dozen. 

12  dozen 1  gross. 

12  gross,  or  144  dozen 1  great  gross. 

20  things 1  score. 

196  pounds  of  flour 1  barrel. 

200  pounds  of  beef  or  pork 1  barrel. 

100  pounds  offish 1  quintal 

18  inches 1  cubit 

22  inches  (nearly,) 1  sacred  cubit. 

Art.  100.   PROMISCUOUS    EXAMPLES. 

1.  What  cost  2bu.  plums,  at  Sets,  a  pt. ?      Ans.$6A0 

2.  3bu.  2pk,  peaches,  at  50cts.  a  pk.  ?         Ans.  $7. 

3.  3  pk.  3  qt.  barley,  at  3  cts.  a  pt.  ?  Ans.  $1 .  62 

4.  At  15  cents  a  peck,  how  many  bushels  of  apples  can 
be  bought  for  ^3  ?  Ans.  5  bu. 

5.  If  salt  cost  2  cents  a  pint,  how  much  can  be  bought 
with  $1.66?  Ans.  Ibu.  1  pk.  Iqt.  Ipt. 

6.  I  put  91  bushels  of  wheat  into  bags  of  3bu.  2pk. 
each  :  how  many  bags  were  required  ?  Ans.  26. 

Reduce  both  quantities  to  pecks,  and  then  divide. 

7.  What  is  the  value  of  1  lb.  3pwt.  of  gold  ore,  at  3 
cents  a  grain?  Ans.  $174.96 

8.  How  many  spoons,  each  weighing  2oz.  5pwt.,  can  be 
made  from  21b.  5oz.  5pwt.  of  silver?  Ans.  13. 

9.  How  many  rings,  weighing  5  pwt.  7  gr.  each,  can  be 
made  from  1  lb.  8  oz.  18  pwt.  Igr.  of  gold?         Ans.  79. 

10.  If  lib  45  of  calomel  be  divided  into  doses  of  15 
grains  each,  and  sold  at  12cts.  5m.  a  dose,  what  will  it 
amount  to  ?  Ans.  $50. 


REDUCTION    OF    COMPOUND  NUMBERS.  101 

What  will  be  the  cost  op 

11.  IR)  15  13  opium,  at4cts.  a  9?      Ans.  $12.60 

12.  6cwt.  Iqr.  raisins,  at  3cts.  a  lb.?     Ans.  $18.75 

13.  IT.  Icwt.  rice,  at  S2.25  a  qr.?  Ans.  $189. 

14.  7  lb.  80Z.  copper,  at  5cts.  an  oz.?  Ans.  $6. 

15.  A  physician  put  up  316  doses  of  rhubarb,  of  20  gr. 
each :  how  much  did  he  use?         Ans.  Itbl^  15  19- 

16.  How  many  nails,  weighing  4  drams  each,  are  in  a 
parcel  weighing  151b.  9oz.  12 dr.?  Ans.  999. 

17.  I  bought  44cwt.  2qr.  21b.  of  cheese;  each  weighed 
91b.  15 oz. :  how  many  cheese  did  I  buy?  Ans.  448, 

18.  How  many  kegs,  of  841b.  each,  can  be  filled  from 
a  hhd.  of  sugar  of  14 cwt.  Iqr.  31b.?  Ans.  17. 

19.  How  many  boxes,  containing  121b.  each,  can  be 
filled  from  7  cwt.  2qr.  61b.  of  tobacco?  Ans.  63. 

20.  If  a  family  use  31b.  13  oz.  of  sugar  in  a  week,  how 
long  will  6 cwt.  101b.  last  them?  Ans.  160 wk. 

21.  What  will  2A.  3R.  5P.  of  land  cost,  at  20  cts.  a 
perch?  Ans.  $89. 

22.  What  cost  2sq.  yd.  2  sq.ft.  of  ground,  at  5  cents  a 
square  inch  ?  Ans.  $144. 

23.  How  many  leaves,  3  in.  long  and  2  in.  wide,  can 
be  cut  from  1  sq.  yd.  of  paper  ?  Ans.  216. 

24.  A  farmer  has  a  field  of  16 A.  IR.  13P.,  to  divide 
into  lots  of  1  A.  1 R.  1  P.  each :  how  many  lots  will  it 
make?  Ans.  13. 

25.  How  many  cu.  in.  in  a  block  of  marble  2  ft.  long, 
2ft. high,  2ft.  wide?  Ans.  13824cu.  in.,  or8cu.  ft. 

26.  One  cubic  foot  of  water  weighs  1000  oz.  Avoirdu- 
pois :  what  do  5cu.  ft.  weigh?  Ans.  3121b.  8oz. 

27.  What  is  the  weight  of  a  quantity  of  water  occupy- 
ing the  space  of  1  cord  of  wood,  each  cu.  ft.  of  water  weigh- 
ing 1000  oz.  Avoirdupois?  Ans.  4T. 

28.  A  cubic  foot  of  oak  weighs  950  oz.  Avoirdupois : 
what  do  2  cords  of  oak  weigh?  Ans.  7T.  12 cwt. 

29.  How  many  pieces,  of  13  yd.  3qr.  2na.  each,  are 
there  in  666  yards  ?  Ans.  48. 

30.  How  many  suits  of  clothes,  each  containing  5  yd. 
Iqr.,  can  be  made  from  147  yards?  Ans.  28. 


102  RAY'S   PRACTICAL    ARITHMETIC. 

31.  How  many  coat  patterns  of  2  yd.  Iqr.  Ina.  each, 
can  be  cut  from  37yd.  of  cloth?  Ans.  16. 

32.  How  many  suits  of  clothes  of  3  yd.  2qr.  each,  can 
be  cut  from  70  Ells  Flemish  of  cloth  ?  Ans.  15. 

33.  Find  the  cost  of  1  hogshead  of  wine,  at  5  cents 
a  gill.  Ans.  8100.80 

34.  Find  the  cost  of  5bl.  of  molasses,  each  containing 
31  gal.  2qt.,  at  10  cents  a  quart.  Ans.  $63. 

35.  At  5  cents  a  pint,  what  quantity  of  molasses  can  be 
bought  for  82  ?  Ans.  5  gal. 

36.  At  let.  5m.  a  gill,  how  much  cider  can  be  bought 
for  812?  ^ns.  25  cral. 

37.  How  many  dozen  bottles,  each  bottle  holding  3qt. 
Ipt.,  can  be  filled  from  Ihhd.  of  cider?         Ans.  6  doz. 

38.  How  many  kegs  of  4gal.  3qt.  Ipt.  each,  can  be 
filled  from  1  wine  hhd.  ?      Ans.  12;  and  4gal.  2qt.  left. 

39.  How  many  bottles,  each  holding  1  gal.  1  qt.  1  pt., 
can  be  filled  from  165  wine  gal.?  Ans.  120. 

40.  What  will  1  hogshead  of  beer  cost,  at  3  cents  a 
quart?  ^ns.  86.48 

41.  The  human  heart  beats  70  times  a  minute :  how 
many  times  will  it  beat  in  a  day?  Ans.  100800. 

42.  How  many  seconds  in  the  month  of  February, 
1840?     (See  Note  1,  page  98.)  Ans.  2505600  sec. 

43.  How  many  hours  longer  is  January  than  February, 
A.  D.  1839?     (See  Note  1,  page  98.)  Ans.  72  hr. 

44.  What  are  the  leap  years  between  1837  and  1850? 

(See  Note  1,  page  98.)        Ans.  1840,  1844,  1848. 

45.  The  exact  length  of  the  solar  year  is  365  da.  5hr. 
48min.  48  sec.  :  how  many  such  years  are  contained 
in  1609403328  seconds  ?  Ans.  51  years. 

46.  How  much  time  will  a  man  gain  in  60 yr.,  of  365 
da.  each,  by  rising  30  min.  earlier  each  day  ? 

Ans.  456  da.  6hr. 

47.  If  a  ship  sail  8  miles  an  hour,  how  many  miles  will 
she  sail  in  3wk.  2da.  3hr.?  Ans.  4440mi. 

48.  If  a  planet  move  through  2°  in  a  day,  how  long 
will  it  require  to  move  through  2=*  4°?  Ans.  32  da. 


ADDITION    OF   COMPOUND   NUMBERS.  103 

49.  In  what  time  will  one  of  the  heavenly  bodies  move 
through  a  quadrant,  (90°),  at  the  rate  of  43'  12''  per 
minute?  Aiis.  2hr.  5min. 

What  will  be  the  cost 

50.  Of  2  reams  paper,  at  20  cts.  a  quire?     Ans.  S8. 

51.  3  quires  paper,  at  2  cts.  a  sheet?         Ans.  $1.44 

52.  3  dozen  apples,  at  2  cts.  each  ?  Ans.  72  cts. 

53.  4  doz.  doz.  oranges,  at  3 cts.  each?  Ans.  $17.28 

54.  5  gross  buttons,  at  5  cts.  a  doz.  ?  Ans.  $3. 

55.  1  bl.  of  flour,  at  4  cts.  a  pound?         Ans.  $7.84 

56.  1  bl.  pork,  at  12 cts.  5  m.  a  pound?      Ans.  $25. 

57.  A  farmer  started  for  market  with  6  dozen  dozen 
eggs ;  he  broke  half  a  doz.  doz.,  and  sold  the  remainder 
at  1  ct.  each  :  what  did  they  amount  to  ?         Ans.  $7.92 

]^^^  For  additional  problems,  see  Ray's  Test  Examples. 

ADDITION  OF  COMPOUND  NUMBERS. 

Art.  101.  The  process  of  uniting  numbers  of  dif- 
ferent denominations  into  one  sum  or  amount,  is  termed 
Addition  of  Compound  Numhers. 

1.  A  farmer  sold  three  lots  of  wheat :  the  first  lot  con- 
tained 25  bu.  3pk.  4qt. ;  the  second,  14  bu.  5qt.  Ipt. ; 
the  third,  32  bu.  1  pk.  1  pt.  :  find  the  amount. 

Solution. — In  writing  the  numbers,  place  units  of  the  same 
denomination  under  each  other,  since  only  numbers  of  the  same 
name  can  be  added.     Art.  20. 

Beginning  with  the  column  of  the  lowest  operation. 

denomination,  and  adding  together  the  units  bu.  pk.  qt.  pt. 
in  it,  the  sum  is  2  pints,  which  is  reduced  to  25340 
quarts  by  dividing  by  2,  the  number  of  pints  14  0  5  1 
in  a  quart,  and  there  being  no  pint  left,  32  1  0  1 
write  a  0  in  the  order  of  pints,  and  carry  ^  ^  I  ~  T 
the  1  (quart)  to  the  column  of  quarts.  "^  ^ 

Then,  adding  the  numbers  in  the  column  of  quarts,  the  sum 
is  10  quarts,  which,  being  reduced,  make  1  peck  and  2  quarts;  write 
the  2  quarts  in  the  column  of  quarts,  and  carry  the  1  (peck)  to 
the  column  of  pecks.     The  number  in  the  column  of  pecks  added, 


104 


RAY'B   PRACTICAL   ARITHMETIC. 


make  5  pk.,  which  reduces  to  1  bushel  and  1  pk. ;  write  the  1  pk. 
in  the  column  of  pecks,  and  carry  the  1  bu.  to  the  column  of 
bushels  :  the  sum  of  the  numbers  in  this  column  is  72,  which  is 
written  in  the  same  manner  as  the  sum  of  the  numbers  in  the  left 
hand  column  in  Simple  Addition. 


bu.    pk.  qt.  pt. 

(2)    3      2  0  1 

4      0  6  1 

13  7  1 

9      2  G  1 


bu.  pk.  qt.  pt. 

(3.)  7  3  7  1 

6  2  0  0 

9  2  4  1 

24  0  4  0 


Hule  for  Addition. — 1.  Write  the  numbers  to  he  added, 
placing  units  of  the  same  denomination  under  each  other. 

2.  Begin  with  the  lowest  order,  add  the  numbers,  and  divide 
their  sum  by  the  number  of  units  of  this  denomination,  which 
make  a  unit  of  the  next  higher.  Write  the  remainder  under  the 
column  added,  and  carry  the  quotient  to  the  next  column. 

3.  Proceed  in  the  same  manner  with  all  the  columns  to  the  last, 
under  which  write  its  entire  sum. 

Proof. — The  same  as  in  Addition  of  Simple  Numbers. 

Note. — In  writing  Compound  Numbers,  if  any  intermediate 
denomination  is  wanting,  supply  its  place  with  a  cipher. 

Rem. — In  adding  Simple  Numbers,  we  carry  one  for  every  ten, 
as  ten  units  of  a  lower  order  make  one  unit  of  the  next  higher. 

But  in  Compound  Numbers,  we  carry  one  for  thefiumber  of  units 
of  each  lower  order,  which  make  one  unit  of  the  next  higher. 


(4.) 


EXAMPLES   IN    ADDITION. 


TROY  WEIGHT. 


APOTHECARIES. 


lb. 

oz. 

pwt. 

gi*- 

13 

0 

17 

19 

42 

11 

19 

23 

31 

9 

0 

4 

(5) 


lb 

s 

0 

9 

gr 

7 

2 

7 

2 

14 

4 

6 

5 

0 

17 

8 

1 

6 

1 

10 

Review. — 101.  What  is  Addition  of  Comiwund  Numbers  ?  Why  place 
units  of  tho  samo  denomination  under  each  other  ?  Where  do  you  begin 
to  add  ?  How  is  tho  addition  performed  ?  Repeat  tho  Rule.  Rem.  For 
what  do  you  carry  ono  ? 


ADDITION   OF    COMPOUND   NUMBERS. 


105 


f 


AVOIRDUPOIS   WEIGHT. 

T.  cwt.  qr.  lb.            cwt.  qr.  lb.  oz.  dr. 

(6)45   3  3  20       (7.)  31  3  17  14  14 

14  14  1  14          25  0  20  10  10 

19  17  2  11          37  2  16  11  13 


LONG   MEASURE. 

mi.   fur. 

(8.)  14     7 
12     3 
11     2 

rd. 
24 
16 
19 

yd.     ft.     in. 

(9.)    2     1      11 
5     2       9 
3     1       8 

A.     R. 

(10.)  41      1 

64     2 

193     3 

SQUARE   MEASURE. 
P. 
11                                    (11. 

24 
35 

CUBIC   MEASURE. 

sq.yd.  sq.ft.  sq.in. 

)  15      8      115 
20      7      109 
14      5      137 

C.     cu.  ft.   cu.in.                            cu.yd.  cu.ft.  cu.in. 

(12.)  13      28     390                    (13)  50      18      900 
15     90     874                             45      17      828 
20     67      983                            46     20     990 

CLOTH   MEASURE. 

yd.  qr.  na. 

4.)  14    3    2 
6    2     1 
5     1     3 

E.Fl.  qr.  na. 

(15.)   1112 
13    3    0 
12     1     1 

WINE   MEASURE. 

E.En.  qr.  na 

(16.)  18     0     3 
16     1     2 
11     3     1 

T.   hhd. 

(17.)  9      1 

7  2 

8  3 

gal.    qt. 

40      1                     (18.) 

58      3 

61      2 

gal.    qt.     pt.    gi. 

40     3      1      3 
16      1      0     2 
71      2      1      2 

TIME 

MEASURE. 

CIRCULAR. 

mon.  wk.da. 

hr.    min.  sec. 

o            /          // 

(19.)  2     3    7 
1     2    4 
3     1    5 

23     51     40            (20.) 
IJ    30    37 
13    27     18 

180      18     20 

101      35     42 

47      11      44 

106 


RAY'S   PRACTICAL  ARITHMETIC. 


21.  Five  loads  of  wheat  measured  thus:  21  bu.  3pk. 
2qt.  Ipt.;  14  bu.  5qt. ;  23  bu.  2pk.  Ipt. ;  18  bu.  Ipk. 
Ipt.;  22 bu.  7qt.  Ipt.:  how  much  in  all?    Ans.  100 bu. 

22.  A  farmer  raised  of  oats  200 bu.  3pk.  Ipt. ;  barley, 
143 bu.  2qt.  Ipt.;  corn,  400 bu.  3pk. ;  wheat,  255 bu. 
Ipk.  5qt.  :  how  much  in  all?  Ans.  1000 bu. 

23.  I  bought  a  silver  dish  weighing  21b.  10  oz.  15pwt. 
21  gr. ;  a  bowl  weighing  1  lb.  1  oz.  16  pwt.  14  gr. ;  a  tank- 
ard, 21b.  8oz.  5 pwt.  12 gr.  :  what  did  all  weigh? 

Ans.  61b.  8oz.  17 pwt.  23 gr. 

24.  A  drusririst  mixed  to";ether  four  articles :  the  first 
weighed  3  5  4  5  19;  the  second,  45  35  29;  the  third, 
45  18 gr.;  and  the  fourth,  6  5  5  5  29  18gr. :  what  was 
the  weight  of  all?  Ans.  1  ft  35  25  16 gr. 

25.  A  grocer  sold   5hhd.  of  sugar  :  the  first  weighed 
8cwt.  Iqr.  111b.;  the  2d,  4cwt.  2qr.  141b.;  the  3d,  5cwt 
191b.;    the  4th,  7cwt.  3qr. ;    the  5th,  7cwt.   3qr.   91b. 
what  did  all  weigh?  Ans.  33cwt.  3qr.  31b. 

26.  Add  together,  131b.  lloz.  15dr. ;  171b.  13ozl 
lldr. ;    141b.    14oz. ;    161b.    lOdr.  ;    191b.   7oz.    12dr. 


and  171b.  9oz.  9dr 


Ans.  991b.  9oz.  9dr 


27.  Two  men  depart  from  the  same  place  :  one  travels 
104 mi.  Ifur.  10 rd.  due  east;  the  other,  95 mi.  6  fur. 
30 rd.  due  west :  how  far  are  they  apart?     Ans.  200 mi. 

28.  A  man  has  3  farms:  in  the  1st  are  186 A.  3R. 
14  P.;  in  the  2d,  286  A.  17  P. ;  in  the  3d,  113  A.  211. 
9 P. :  how  much  in  all?  Ans.  586 A.  211. 

29.  Add  together,  17sq.yd.  3sq.ft.  119sq.in.  ;  18 
sq.yd.  141sq.in.  ;  23sq.yd.  7sq.ft.  ;  29sq.yd.  5sq.ft. 
116sq.in.  Aus.  88 sq.yd.  8 sq.ft.  88sq. in. 

30.  A  has  4  piles  of  wood:  in  the  first  7  C.  78cu.ft. ;  the 
2d,  16C.  24cu.ft.  ;  the  3d,  35C.  127cu.ft. ;  the4th,29C. 
lOcu.ft.  :  how  much  in  all?  Ans.  88C.  lllcu.ft. 

31.  Bought  5  pieces  of  cloth  ;  the  first  contained  17yd. 
3qr.  2na.  ;"the  2d,  13yd.  2qr.  Ina.;  the  3d,  23yd.  2na.; 
the  4th,  27yd.  Iqr.  2na.  ;  the  5th,  29yd.  Iqr.  2na.: 
find  the  amount.  Aus.  111yd.  Iqr.  Ina. 


SUBTRACTION   OF    COMPOUND   NUMBERS.         107 

32.  I  sold  to  A,  73hhd.  43o;al.  3qt.  Ipt.  of  wine  ;  to 
B,  27hhd.  3gal. ;  to  C,  15hhd.  3qt.  Ipt.;  to  D,  162 hhd. 
Iqt. :  how  much  in  all?  Ans.  277 hhd.  48 gal. 

33.  I  sold  beer  as  follows:  Ibl.  28  gal.  ;  Ibl.  17  gal. ; 
5bl.  2gal.;  Igal.  2qt.  Ipt.;  7gal.  2qt.  Ipt.;  18gal. 
3qt. ;    and  33gal.  :  what  is  the  quantity?       Ans.  lObl. 

34.  What  day  of  the  year  is  the  1st  of  May  in  common 
years?  Ans.  121st. 

35.  What  the  4th  of  July  in  leap  years?    Ans.  186th. 

36.  Through  how  many  degrees  does  a  ship  pass,  in 
sailing  from  Cape  Horn,  latitude  55°  58'  30"  south,  to 
New  York,  lat.  40°  42'  40"  north?        Ans.  96°  41'  10". 


SUBTRACTION    OF    COMPOUND    NUMBERS. 

Art.  102.  The  process  of  finding  the  difference  between 
two  numbers  of  different  denominations,  is  termed  Sub- 
traction of  Compound  Numbers. 

1.  I  have  67bu.  Ipk.  3qt.  of  wheat:  how  much  will 
remain  after  selling  34 bu.  2pk.  Iqt.? 

Solution. — In  writing  the  numbers,  the  less  operatiox. 

is  placed  under  the  greater,  for  convenience.  bu.    pk.    qt. 

Place  units  of  the  same  denomination  under  67  1  3 
each  other,  because  a  number  can  be  subtracted  .34  2  1 
only  from  one  of  the  same  name.  09      q      0 

Then    subtract   1    (quart)    from    3    (quarts), 
which  leaves  2  (quarts)  to  be  placed  in  the  column  of  quarts. 

In  the  next  column,  2  (pecks)  can  not  be  taken  from  1  (peck), 
but  we  can  take  1  bushel  from  67  bushels,  reduce  it  to  pecks,  and 
add  them  to  the  1  peck  ;  this  gives  5  pecks,  from  which  take  2 
pecks,  and  3  pecks  are  left,  to  be  placed  in  the  column  of  pecks. 

Then  take  34  bushels  from  66  bushels,  as  in  Simple  Subtraction, 
and  write  the  remainder,  32  bushels,  in  the  column  of  bushels. 

Instead  of  diminishing  the  upper  number  (67  bu.)  by  1,  the  result 
will  be  the  same  to  increase  the  lower  number  (34  bu.)  by  1.  Art.  26.^ 

Review. — 102.  "What  is  Subtraction  of  Compound  Numbers  ?  In 
writing  the  numbers,  why  place  the  less  under  the  greater  ? 


108 


RAY'S   PRACTICAL   ARITHMETIC. 


bu.  pk.    qt.    pf. 

(2.)  From  12  0      1      0 

Take     8  2      11 

Ans.     3  17      1 


bu.    pk.  qt.  pt, 

(3.)  5      0  0  0 

10  0      1 

3     3  7      1 


Rule  for  Subtraction. — 1.  Write  the  numbers,  the  less  under 
the  greater,  placing  units  of  the  same  denomination  under  each 
other. 

2.  Begin  with  the  lowest  order,  and,  if  possible,  take  the  lower 
number  from  the  one  above  it,  as  in  Simple  Subtraction. 

3.  But,  if  the  lower  number  of  any  order  be  greater  than  the 
upper,  increase  the  upper  number  by  as  many  units  of  that  de- 
nomination as  make  one  of  the  next  higher ;  subtract  as  before, 
and  carry  one  to  the  lower  number  of  the  next  higher  order. 
Proceed  in  the  same  manner  with  each  denomination. 

Proof. — As  in  Subtraction  of  Simple  Numbers. 

Rem. — In  Simple  Subtraction,  when  any  lower  figure  is  greater 
than  the  upper,  we  borrow  ten,  ten  units  of  a  lower  order  making  a 
unit  of  the  next  higher.  In  Compound  Numbers,  when  the  lower 
number  of  any  order  is  greater  than  that  above  it,  borrow  thenuni' 
her  of  units  in  that  order  which  makes  a  unit  of  the  next  higher. 


TROY   WEIGHT. 

lb.    oz.    pwt.    gr. 

(4.)  From  18     6     IG     13 
Take    9     6     18      6 


T. 

(5.)  14 
10 


AVOIRDUPOIS. 

cwt.  qr.     lb.     oz,  dr. 

12    2    20     11  14 

9    3     14     12  11 


LONG   MEASURE. 


mi.    fur.     rd. 

(G.)  From  18      5      36 

Take  11      4      38 


(7.) 


yd.   ft. 

4      1 
2      1 


in. 

10 
11 


A.     R. 

(8.)  From  327    3 
Take    77     2 


SQUARE   MEASURE. 

P.                                   sq.yd.  sq.ft.  sq.in. 

28                          (9.)  19        ()  72 

30                               16        6  112 


TIME   MEASURE. 

da.         hr.     min.  sec. 

(10.)  From  245      17      40  37 

Take  190      11      44  42 


CIRCULAR, 
o  /  // 

(11.)  29      54      53 
19      54     57 


SUBTRACTION   OF    COMPOUND  NUMBERS.         109 

12.  If  2bu.  Ipk.  Iqt.  be  taken  from  a  bag  of  4bu,, 
what  quantity  will  remain?  Ans.  1  bu.  2pk.  7qt. 

13.  From  100  bushels,  take  24  bu.  Ipt. 

Ans.  75 bu.  3pk.  7qt.  Ipt. 

11.  From  191b.  9oz.  of  silver,  take  91b.  9oz.  lOpwt. 
lOgr.  Ans.  91b.  11  oz.  9pwt.  11  gr. 

15.  A  silver  bar  of  81b.  2oz.  llpwt.,  is  divided  into 
2  parts;  the  less  weighs  21b.  4:0Z.  7pwt.  16gr.  :  find  the 
weight  of  the  greater.  Ans.  51b.  10  oz.  Spwt.  8gr. 

16.  From  3tb  3S  15  19  12gr.,  take  lib  7  5  29 
ISgr.  Ans.  lib  85  19  Hgr. 

17.  I  boutrht  461b.  9oz.  of  rice:  after  sellinor  191b. 
4 dr.,  how  much  remained?  Ans.  271b.  8oz.  12 dr. 

18.  A  wagon  loaded  with  hay  weighs  32cwt.  2qr. 
161b.  ;  the  wagon  alone  weighs  8cwt.  2qr.  171b.  :  what 
is  the  weight  of  the  hay?  Ans.  23cwt.  3qr.  241b. 

19.  It  is  about  25000  miles  round  the  earth  :  after  a 
man  has  traveled  100  mi.  Ifur.  Ird.,  what  distance  will 
remain?  Ans.  24899 mi.  6 fur.  39 rd. 

20.  I  had  146  A.  2R.  of  land.  I  gave  my  son  86  A. 
2  R.  14  P. :  how  much  was  left  ?       Ans.  59  A.  3  R.  26  P. 

21.  From  8C.  50cu.ft.  of  wood,  there  is  taken  3C. 
75cu.ft. :  how  much  is  left?  Ans.  4C.  103cu.ft. 

22.  From  25E.F1.,  take  14E.F1.  Iqr.  3na. 

Ans.  lOE.Fl.  Iqr.  Ina. 

23.  From  11  yards  of  cloth,  there  is  cut  3  yd.  2qr. 
2na.  :  what  remains  ?  Ans.  7yd.  Iqr.  2na. 

24.  A  hhd.  of  wine  leaked;  only  51  gal.  Iqt.  2gi.  re- 
mained :  how  much  was  lost  ?    Ans.  11  gah  2  qt.  1  pt.  2gi. 

25.  From  5da.  lOhr.  27min.  15sec.,  take  2da.  4hr. 
13min.  29 sec.  Ans.  3 da.  6hr.  13min.  46 sec. 


Eevietv. — 102.  "Why  place  units  of  the  same  denomination  under  each 
other  ?  At  which  column  begin  to  subtract  ?  "Why  ?  How  is  the  sub- 
traction performed  ?     Repeat  the  Rule.     What  is  the  proof? 

102.  Rem.  In  subtraction  of  Compound  Numbers  when  the  lower  num- 
ber of  any  order  is  greater  than  the  upper,  what  is  to  be  done  ? 


110  RAY'S    PRACTICAL    ARITHMETIC. 

Art.  103.  To  find  the  Time  between  any  Two  Dates. 

Subtract  the  first  date  from  the  last,  numbering  the  mouths 
according  to  their  order.     See  note  3,  page  98. 

Note. — In  finding  the  time  between  two  dates,  and  also  in 
Interest,  consider  30  days  1  month,  and  12  months  1  year. 

26.  A  note,  dated  April  14th,  1835,  was  paid  Feb- 
ruary 12th,  1837  :  find  the  time  between  these  dates. 

Solution. — In  writing  the  dates,  observe  operation. 

that  February  is  the  2d  month,  and  April,  yr.      mon.    da. 

the  4th.     In  subtracting,  14  days  can  not  1837      2      12 

be  taken  from  12;  therefore  add  30  to  the  12  1835      4      1-4 

days,  subtract,  and  carry  1  to  the  4  months.  ^       q      ^^ 
As5  months  can  not  be  taken  from  2  months, 
add  12  to  the  latter,  subtract,  and  carry  1  to  the  years. 

27.  The  Independence  of  the  United  States  was  declared 
July  4th,  177(3  :  what  length  of  time  had  elapsed  on  the 
5th  of  March,  1857?  Ans.  80yr.  8mon.  Ida. 

28.  A  certain  man  was  born  June  24th,  1822 :  what 
was  his  age  Aug.  1st,  1848?         Ans.  26 yr.  Imon.  7 da. 

29.  A  man  was  born  Nov.  25th,  1807  ;  his  son  was 
born  June  28th,  1832  :  what  is  the  difference  of  their 
ages?  Ans.  24 yr.  7 mon.  3 da. 

30.  The  latitude  of  the  Cape  of  Good  Hope  is  33°  55' 
15"  south;  that  of  Cape  Horn,  55°  58'  30"  south  :  find 
the  difference  of  latitude.  Ans.  22°  3'  15". 

31.  The  latitude  of  Gibraltar  is  36°  6'  30"  north  ;  that 
of  North  Cape,  in  Lapland,  71°  10'  north:  what  their 
difference  of  latitude  ?  Ans.  35°  3'  30". 

32.  A  ship  departs  from  latitude  10°  25'  48"  north, 
and  sails  north  to  latitude  50°  :  through  how  many  de- 
grees of  latitude  has  she  sailed?  Ans.  39°  34'  12". 

33.  Take  5  quires  11  sheets  from  a  ream  of  paper,  and 
how  much  will  remain?  Ans.  14  quires  13  sheets. 

34.  A  man  started  to  market  with  6  doz.  doz.  eggs  :  he 
broke  half  a  doz.  doz. :  how  many  were  left?      Ans.  792. 


OPERATION. 

bu. 

pk.    qt. 

2 

3       4  amt.  in  1  bag. 

5  number  of  bags 

COMPOUND  NUMBERS.  Ill 


MULTIPLICATION    OF    COMPOUND    NUMBERS. 

Art.  104.  The  process  of  taking  a  number  consisting 
of  different  denominations,  a  certain  number  of  times,  is 
termed  MuUipUcatioii  of  Compound  Numbers. 

1.  A  farmer  takes  to  mill  5  bags  of  wheat,  each  con- 
taining 2bu.  3pk.  4qt. :  how  much  in  all? 

Solution, — Begin  at  the 
lowest  denomination:  5  times 
4  quarts  are  20  quarts,  which 
reduced,  make  2  pecks  and  4 
quarts;  write  the  4  (quarts) 
in  the  column  of  quarts,  and  14       1       4  amt.  in  5  bags, 

carry  the  2  pecks. 

Next,  multiply  the  3  pk,  by  5,  making  15  pk.,  to  which  add  the 
2  pk.  carried,  making  17  pk.,  which,  reduced,  give  4  bu.  and  1  pk.  ; 
write  the  1  (pk.)  in  the  column  of  pk.,  and  carry  the  4  bu. 

Then,  multiply  the  2  bu.  by  5,  add  to  the  product  the  4  bu.  carried, 
and  we  have  14  bu.  to  be  written  in  the  column  of  bu. 

In  Compound,  as  in  Simple  Multiplication,  multiply  the  lowest 
denomination  first,  so  as  to  carry  from  a  loioer  to  a  higher  order. 

2.  Multiply  2  bu.  5  qt.   1  pt.  by  6.      Arts.  13  bu.  1  qt. 

3.  Multiply  2  bu.  2  pk.  2  qt.  by  9,      Ans.  23  bu.  2  qt. 

Rule  for  Multiplication. — 1.  Write  the  multiplier  under 
the  loicest  denomination  of  the  multiplicand. 

2.  Multiply  the  lowest  denomination  first,  and  divide  the 
product  by  the  number  of  units  of  this  denomination,  which 
make  a  unit  of  the  next  higher ;  write  the  remainder  under  the 
denomination  multiplied,  and  carry  the  quotient  to  the  product 
of  the  next  higher  denomination. 

Review. — 103.  How  find  the  time  between  any  two  dates  ?  104,  What 
is  Multiplication  of  Compound  Numbers  ?  Why  multiply  >,he  lowest  de- 
nomination first  ?     Repeat  the  Rule. 


112  RAY'S   PRACTICAL   ARITHMETIC. 

3.  Proceed  in  like  manner  with  all  the  denominations,  writing 
the  entire  product  at  the  last. 

Proof. — The  same  as  in  Simple  Multiplication. 

Rem. — 1.  In  Simple  Multiplication,  we  carry  one  for  every  ten, 
because  ten  uuits  of  a  lower  order  make  one  unit  of  the  next  higher. 

In  Compound  Multiplication,  we  carry  one  for  the  number  of  unit? 
in  each  lower  order  which  make  a  unit  of  the  next  higher. 

2.  The  multiplier  is  always  an  abstract  number,  and  shows  how 
many  times  the  multiplicand  is  to  be  taken. 

4.  If  4bu.  3pk.  3qt.  Ipt.  of  wheat  make  Ibl.  of  flour, 
how  much  will  make  12  bl.  ?  Ans.  58 bu.  Ipk.  2qt. 

5.  What  is  the  weight  of  6  silver  spoons,  each  weigh- 
ing 2oz.  llpwt.  6gr.  ?  Ans.  lib.  3oz.  7pwt.  12gr. 

6.  What  is  the  weight  of  10  bars  of  silver,  each  10  oz. 
lOpwt.  10 gr.  ?  Ans.  81b.  9oz.  4pwt.  4gr. 

7.  I  put  up  8  packages  of  medicine,  of  4  5  29  15  gr. 
each:  what  did  all  weigh?  Ans.  2 it)  8  5  7  5  19- 

8.  Find  the  weight  of  9hhd.  of  sugar,  of  8cwt.  2qr. 
121b.  each.  Ans.  3T.  17cwt.  2qr.  81b. 

9.  How  much  hay  in  7  loads,  each  weighing  lOcwt. 
3qr.  141b.?  Am.  3T.  16cwt.  231b. 

10.  If   a  ship  sail  208  mi.  4 fur.  16rd.  a  day,  how  far 
will  she  sail  in  15  days?  Ans.  3128 mi.  2 fur. 

11.  If  a  man  travel  30 mi.  4fur.  10 rd.  a  day,  how  far 
will  he  travel  in  12da.  ?  Ans.  366mi.  3fur. 

12.  Multiply  130  A.  3R.  30  P.  by  4.     Ans.  523  A.  311. 

13.  Multiply  23cu. yd.  9cu.ft.  228cu.in.  by  12. 

^«s.  280cu.yd.  Icu.ft.  lOOScu.in. 

14.  How  many  yards  in  6  pieces  of  muslin,  of  26  yd. 
2qr.  2na.  each?  Ans.  159yd.  3qr. 

15.  Multiply  62 gal.  Iqt.  Ipt.  by  8.         Ans.  499gal. 

16.  How   many   gallons   iu    5    casks,    each    containing 
123gal.  2qt.  Ipt?  Ans.  618gal.  Ipt. 


Review. — 104.  Rem.  1.  In  Compound  Multiplication,  for  what  number 
do  you  carry  one?    2.  Id  tho  multiplier  concrete  or  abstract? 


MULTIPLICATION    OF    COMPOUND    NUMBERS.     113 

17.  In  a   solar   year   are    365  da.  5hr.  48min.  48  sec. : 
how  long  has  B  lived,  who  is  12  years  old? 

Ans.  4382  da.  211ir.  45min.  36  sec. 

18.  Multiply  4°  11'  15"  by  8.  An^.  V  3°  30'. 

19.  How  many  buttons  in  3  great  gross?     Ans.  5184. 

20.  How    much    water    in    48    casks,    each   containing 
62  gal.  Iqt.  Ipt.  Igi.?     Here,  48  =  6  X  8.   (See  Art.  47.) 


gal. 

qt. 

pt. 

gi 

62 

1 

1 

1 

§ 

6 

1— 

374 

1 

1 

2 

g 

8 

o 

water  in  6  casks. 

2995       2       U  0  =  water  in  48  casks. 

21.  Multiply  2bu.  3pk.  5qt.  by  24.    ^?is.  69bu.  3pk. 

22.  3mi.  5fur.  16rd.  X  60.  Ans.  220mi.  4fur. 

23.  6  A.  3R.  30  P.  X  56.  Ans.  388  A.  2R. 

24.  8cwt.  2qr.  141b.  12  oz.  13  dr.  X  22. 

Ans.  9T.  lOcwt.  Iqr.  9oz.  14 dr. 

25.  3gal.  2qt.  Ipt.  Igi.  X  112.     (112  =  8X2X7.) 

Ans.  6hhd.  31  gal.  2qt. 

Note. — When  the  multiplier  exceeds  12,  and  is  not  a  composite 
number,  it  is  most  convenient  to  multiply  each  denomination,  and 
reduce  it  separately,  and  write  only  the  result. 

26.  Multiply  16cwt.  2qr.  241b.  by  119. 

Ans.  99  T.  12cwt.  61b. 

27.  37  yd.  3qr.  2na.X89.     Ans.  3370  yd.  3qr.  2na. 

28.  47  gal.  3  qt.  lpt.X59. 

Ans.  44hhd.  52 gal.  2qt.  Ipt, 

29.  A  travels  27  mi.  3 fur.  35  rd.  in  1  day  :  how  far  will 
he  travel  in  Imon.  of  31  days?  Ans.  852 mi.  5rd. 

30.  In  17  piles,  each  containing  7C.  98cu.  ft.  :    what 
quantity  of  wood?  Ans.  132 C.  2cu.  ft. 

3d  Bk.  8 


114  RAY'S   PRACTICAL    ARITHMETIC. 

DIVISION    OF    COMPOUND    NUMBERS. 

Art.  105.  The  process  of  dividing  numbers  consisting 
of  different  denominations,  is  termed  Division  of  Com- 
pound Numhers. 

The  Divisor  may  be  either  a  Simple,  or  Compound 
Number.     This  gives  rise  to  two  cases  : 

First  Case. — To  find  how  often  one  Compound  Number  is 
contained  in  another  Compound  Number. 

Second  Case. — To  divide  a  compound  number  into  a  given 
number  of  equal  parts.     (See  example  1,  below.) 

Note. — Examples  of  the  First  Case  are  solved  by  reducing  both 
divisor  and  dividend  to  the  same  denomination,  and  then  dividing. 
They  are  treated  of  under  Reduction.    See  examples  6,  8,  page  100. 

Examples  of  the  Second  Case,  are  usually  considered  under  the 
head  of  Compound  Division. 

Art.  106.  1.  Divide  14 bu.  Ipk.  7qt.  Ipt.  of  wheat 
equally  among  3  persons. 

Solution. — Divide  the  highest  denom-  operation. 

ination  first,  as  in  Simple  Numbers,  that  bu.    pk.   qt.   pt. 

if  there  be  a  remainder,  it  may  be  reduced        3)14      1      7      1 

to  the  next  lower  order,  and  added  to  it.  ,         \      7.      Tt      7 

^.    , .  .  •      ,  ^  .•  J      -4»s.  4     d     J      1 

First,  8m  14  is  contained  4  times,  and 

2  bushels  left;  write  the  4  in  the  order  of  bushels,  and  reduce  the 

remaining  2  bushels  to  pecks,  to  which  add  the  1  in  the  order  of 

pecks,  and  the  sum  is  9  pecks,  which,  divided  by  3,  gives  a  quotient 

of  3  pecks,  to  be  written  in  the  order  of  pecks. 

Next,  divide  7  quarts  by  3,  and  the  quotient  is  2  quarts,  with 

1  quart  remainder;  write   the  2   quarts   in   the  order  of  quarts; 

reduce  the  1  quart  to  pints,  and  add  it  to  the  1  in  the  order  of 

pints;  the  sum  is  3  pints,  and  this,  divided  by  3,  gives  a  quotient 

of  1  pint,  which  write  in  the  order  of  pints. 


(2.)       bu.    pk.    qt,  (3.)      da.      hr.     min 


sec. 


7)33     2     G  5)17     12     56     15 


Ans.  4     3     2  Ans.  3     12     11     15 


DIVISION   OF   COMPOUND   NUMBERS.  115 

Rule  for  Division. — 1.  Write  the  quantity  to  he  divided  in 
the  order  of  its  denominations,  beginning  with  the  highest; 
write  the  divisor  on  the  left. 

2.  Begin  with  the  highest  denomination^  divide  each  number 
separately^  and  write  the  quotient  beneath. 

S.  If  a  remainder  occurs  after  any  division,  reduce  it  to  the 
next  loiver  denomination,  and,  before  dividing,  add  to  it  the 
numbei-  of  its  denomination. 

Proof. — The  same  as  in  Simple  Division. 

Rem. — In  Simple  Numbers  when  a  remainder  occurs  in  dividing 
any  order  except  the  lowest,  it  is  prefixed  to  the  figure  in  the  next 
lower  Older,  which  is  equivalent  to  multiplying  it  by  10,  and  add- 
ing to  the  product  the  figure  in  the  next  lower  order. 

Hence,  in  both  Simple  and  Compound  Division,  each  remainder 
is  multiplied  by  that  number  of  units  of  the  next  lower  order 
which  make  a  unit  of  the  same  order  as  the  remainder. 

Each  partial  quotient  is  of  the  same  denomination  as  that  part 
t  of  the  dividend  from  which  it  is  derived. 

4.  Divide  67  bu.  3pk.  4qt.  Ipt.  by  5. 

Ans.  13 bu.  2pk.  2qt.  Ipt. 

5.  Eight  silver  tankards  of  the  same  size,  weigh  141b. 
8oz.  16pwt.  16 gr. :  what  is  the  weight  of  each? 

Ans.  1  lb.  10 oz.  2pwt.  2gr. 

6.  What  will  1  dollar  weigh ;  the  weight  of  10  dollars 
being  8oz.  12pwt.  12 gr.?  Ans.  17pwt.  6gr. 

7.  Eleven  bl.  of  sugar  weigh  35cwt.   Iqr.  171b.  3oz. 
!7dr. :  find  the  weight  of  one.        An.<i.  3cwt.  221b.  5  dr. 

8.  I  traveled  39 mi.  7 fur.  8rd.  in  7  hours:  at  what 
-rate  per  hour  did  I  travel?  Ans.  5 mi.  5 fur.  24 rd. 

9.  Divide  62yd.  3na.  by  5.         Ans.  12yd.  Iqr.  3na. 

Review. — 105.  What  is  Division  of  Compound  Numbers  ?  What  may 
ithe  divisor  be  ?  What  is  the  first  Case  ?  What  the  second  ?  Note.  How 
I  are  examples  of  the  first  Case  solved  ? 

106.  In  dividing  a  Compound  Number,  why  divide  the  highest  denomi- 
nation first  ?     When  a  remainder  occurs,  how  proceed,  Rule  ? 


116  RAY'S   PRACTICAL   ARITHMETIC. 

Note. — If  the  divisor  exceeds  12,  and  is  a  composite  number, 
we  may  divide  by  its  factors  in  succession,  as  in  Art.  47. 

10.  Divide  69A.  IR.  24P.  by  16. 

(16  =  8X2,  or  4  X  4.)  operation. 

A.     R.     P. 
The  true  remainder  is  found  in  J)d9     1     ^4: 

the  same  manner  as  in  Division  of  8)34     2     32 

iSimple  Numbers.  (Art.  47).  

Ans.  4     1      14 

11.  490bu.  2pk.  4qt. -^100.     Ans.  4bu.  3pk.  5qt. 

12.  2661b.  9oz.  lOdr.  ^50.      ^7is.  51b.   5oz.  5dr. 

13.  3391b.  7oz.  9pwt.  18gr.--42. 

Ans.  81b.  loz.  17  gr. 

14.  114da.  22hr.  45min.  35sec.-v-54. 

Ans.  2 da.  3hr.  5min.  17 sec,  and  17  sec.  rem. 

15.  45T.  18cwt.--17. 

T.     cwt. 

OPERATION.    17)45   18(2 T.  14 cwt.   Ans. 
34 


When  the  divisor 
exceeds  12,  and  is  not 
a  composite  number, 
divide  each  denomi- 
nation as  in  Long 
Division.     Art.  46. 


11 
20 

=  cwt.  in  11 
=  cwt.  to  be  { 

14  cwt. 

220  = 

18  = 

tuns, 
idded. 

17 

)238( 
17 

68 
68 

16.  10271b.  loz.  8dr.--23.       Ans.  441b.  lOoz.  8dr. 

17.  171b.  7oz.  6pwt.  6gr.-=-245.      Ans.  17pwt.  6gr. 

18.  309bu.  2pk.  2qt.-^78.  Ans.  3bu.  3pk.  7qt. 

19.  788mi.  4f'ur.  9rd.--319.       Ans.  2mi.  3fur.31rd. 

Review. — lOH.  Repeat  the  Rule  for  Division.  Rem.  In  both  Simple 
and  Compound  Division,  by  what  is  each  romaindcrinultiplicd  ?  Of  what 
'Icnomi  nation  or  order  id  each  partial  quotient  figure? 


division  of  compound  numbers.         h? 

Art.  107.  promiscuous    examples. 

1.  If  from  lib  of  Ipecac  there  be  taken,  at  one  time 
45  25  13gr.,  and  at  another  85  15  23  14gr.,  how 
much  will  be  left?  Ans.  4  5  33  29  13gr. 

2.  A  silversmith  has  3  pieces  of  silver,  the  first  weigh- 
ing 80Z.  lOpwt.  12gr. ;  the  2d,  9oz.  3pwt.  5gr. ;  the  3d, 
80Z.  9pwt.  7gr.  If  the  loss  in  refining  be  5pwt.  12gr., 
and  the  rest  be  made  into  15  spoons  of  equal  weight,  what 
will  each  spoon  weigh?  Ans.  loz.  14pwt.  12gr. 

3.  I  have  two  farms,  one  104  A.  2R.  37  P.,  the  other, 
87  A.  IR.  38P. :  I  reserve  for  myself  40  A.  IR.,  and 
divide  the  remainder  equally  among  my  3  sons  :  required 
the  share  of  each.  Ans.  50  A.  2R.  25  P. 

4.  A  boy  residing  3 fur.  25  rd.  from  school,  attends 
twice  a  day  :  how  far  does  he  travel  in  30  days  ? 

Ans.  54  mi.  3 fur. 


f 


5.  B  loaned  A  money  on  the  27th  of  June,  1843,  and 
A  paid  it  February  3d,  1845  ;  A  then  lent  B  a  sum  to  be 
kept  5  times  as  long :  how  long  is  B  to  keep  A's  money  ? 

Ans.  8  yr. 

6.  A  ship  in  latitude  35°  30'  north,  sails  20°  35' 
south  ;  then  14°  20'  north;  then  25°  4'  30"  south  ;  then 
6°  19'  20"  north :  what  is  now  her  latitude? 

^KS.  10°  29' 50"  north. 

Art.  108.  LONGITUDE    AND    TIME. 

Difference  of  Longitude  and  Time  between  different  places. 

The  circumference  of  the  earth,  like  other  circles,  is 
divided  into  360°,  (equal  parts),  called  degrees  of  Lon- 
gitude. 

The  sun  appears  to  pass  entirely  round  the  earth  (360°) 
once  in  24  hours,  one  day;  and  in  1  hour,  over  15°. 
(360-^-24=15°).     And, 

As  15°  equal  900^  of  a  degree,  and  1  hour  equals  60 
minutes  of  time,  therefore,  the  sun  passes  in  1  min.  of  time  over 
1 5'  of  a  degree.     (900^-5-60=1  50-     And, 


118  RAY'S   PRACTICAL    ARITHMETIC. 

As  IS"*  equal  900^^  of  a  degree,  and  1  min.  of  time  equals 
60  sec.  of  time,  therefore,  in  1  sec.  of  time  the  sun  passes  over 
1  y'  of  a  degree.     (OOO^^H-  60^1  b'').     Hence  the 

TABLE   FOR   COMPARING   LONGITUDE   AND   TIME. 

15°  of  longitude,  ==  1  hour  of  time. 
IS**  of  longitude,  =  1  min.  of  time. 
1 5''^  of  longitude,  =  1  sec.    of  time. 

1.  How  many  hr.  min.  and  sec.  of  time  correspond 
to  18°  25'  30"  of  longitude?       Ans.  Ihr.  13min.  42sec. 

Analysis. — By  inspection  of  the  Table,  it  is  evident  that, 
Degrees  (°)  of  longitude  divided  by  15,  give  hours       of  time : 
Minutes  (^)  of  longitude  divided  by  15,  give  minutes  of  time: 
Seconds  {^^)  of  longitude  divided  by  15,  give  seconds  of  time. 
Hence,  if  18°  25^  30^^  of  Ion.  be  divided  by  15,  the  quotient  will 
be  the  time  in  hr.  min.  and  sec.  corresponding  to  that  longitude. 

To  find  the  time  corresponding  to  any  diflference  of 
longitude : 

Bule. — Divide  the  longitude  by  15,  according  to  the  Rule 
for  Division  of  Compound  Numbers,  and  mark  the  quotient 
hr.  min.  sec,  instead  of  °   ^  ^\ 

Conversely :  To  find  the  longitude  corresponding  to 
any  difference  of  time  : 

Rule. — Multiply  the  time  by  15,  according  to  the  Rule  for 
the  Multiplication  of  Compound  Numbers,  and  mark  the 
product  °  ^  ^^  instead  of  hr.  min.  sec. 

For  short  methods  of  operation,  see  "  Ray's  Jligher  Arithmetic." 

2.  The  difference  of  longitude  between  two  places 
is  30°  :  what  is  their  diff.  of  time?  Ans.  2hr. 

3.  The  diff.  of  Ion.  between  two  places  is  71°  4':  what 
the  diff.  of  time?  Ans.  4hr.  44min.  IGsec. 

Review. — 108.  At  what  rate  does  the  sun  appear  to  move  in  a  day? 
In  an  hour  ?  In  a  minute?  In  a  second  ?  What  do  degrees,  minutes,  and 
seconds  of  longitude  divided  by  15  give  ?     Repeat  the  Rules. 


DIVISION   OF   COMPOUND  NUMBERS.  119 

p-    4.  The  diff.  of  Ion.  between  New  York  and  Cincinnati 
is  10°  35'  :  what  the  diff.  of  time?      Ans.  42min.  20sec. 

5.  The  diff.  of  time  between  Cincinnati  and  Philadelphia 
is  37min.  20sec.  :  what  the  diff.  of  Ion.  ?        Ans.  9°  20'. 

P"     6.  The  diff.  of  time  between  New  York  and  St.  Louis 
is  Ihr.  "Amin.  56  sec.  :  what  the  diff.  of  Ion.  ?    Ans.  16°  14'. 

7.  The  diff.  of  time  between  London  and  Washington 
is  5hr.  8min.  4sec. :  what  the  diff.  of  Ion.  ?    Ans.  77°  1'. 

r  DIFFERENCE   IN   TIME. 

Art.  109.  It  is  noon  (12  o'clock),  at  any  place  when 
the  sun  is  on  the  meridian  of  that  place ;  and, 

As  the  sun  appears  to  travel  from  the  east  toward  the 
west,  when  it  is  noon  at  any  place,  it  is  aftei'  noon  east  of 
that  pFace,  and  before  noon  icest  of  that  place  : 

Hence,  a  place  has  later  or  earlier  time  than  another, 
according  as  it  is  east  or  west  of  it.     Therefore, 

When  the  time  at  one  place  is  given,  the  time  at  another, 
EAST  of  this,  is  found  hy  ADDING  their  difference  of  time : 
Or,  if  WEST,  hy  SUBTRACTING  their  difference  of  time. 

8.  When  it  is  noon  at  Cincinnati,  what  is  the  time  at 
Philadelphia?  Ans.  37min.  20 sec.  past  noon. 

9.  When  it  is  11  o'clock  A.  M.  at  New  York,  what  is 
the  time  in  Ion.  30°  east  of  New  York  ?       Ans.  1  P.  M. 

W  10.  When  12  o'clock  (noon)  at  Philadelphia,  what  is  the 
time  at  Cincinnati?         Ans.  llhr.  22min.  40 sec.  A.M. 

11.  When  it  is  11  o'clock  A.  M.  at  New  York,  what  is 
the  time  at  St.  Louis?        Ans.  9hr.  55min.  4sec.  A.  M. 

12.  Wheeling,  Ya.,  is  in  Ion.  80°  42'  west:  the  mouth 
of  the  Columbia  river  in  Ion.  124°  west :  when  it  is  1  o'clock, 
P.  M.,  at  Wheeling,  what  is  the  time  at  the  mouth  of 
Columbia  river  ?  Ans.  lOhr.  6min.  48 sec.  A.  M. 

Eeview. — 109.  "When  is  it  noon  at  anyplace?  What  the  time  east 
or  west  of  that  place  ?  "Why  is  the  time  later  east  ?  Why  earlier  west  ? 
Having  the  time  at  one  place,  how  find  the  time  at  another  ? 


120  RAY'S   PRACTICAL   ARITHMETIC. 

VIII.    FACTORING. 

Art.  110.  Definitions. — 1.  An  integer  is  a  whole 
number ;     as,   1,  2,  3,  &c. 

Def.  2.  Whole  numbers  are  divided  into  two  classes  ; 
prime  numbers,  and  composite  numbers. 

Def.  3.  A  prime  number  can  be  exactly  divided  only 
by  itself  and  unity,  (1). 

Thus,     1,  2,  3,  5,  7,  11,  &c.,  are  prime. 

Def.  4.  A  composite  number  (x\rt.  33)  can  be  exactly 
divided  by  some  other  number  besides  itself  and  unity. 

Thus,     4,  6,  8,  9,  10,  &c.,  are  composite. 

Def.  5.  Two  numbers  are  prime  to  each  other  when 
unity  (1),  is  the  only  number  that  will  exactly  ^divide 
both.     Thus,  4  and  5  are  prime  to  each  other. 

Rem. — Two  prime  numbers  are  always  necessarily  prime  to  each 
other.  Also,  two  composite  numbers  are  sometimes  prime  to  each 
other :  thus,  4  and  9  are  prime  to  each  other. 

Def.  6.  An  even  number  can  be  divided  by  2  without  a 
remainder.     Thus,  2,  4,  6,   8,  &c.,  are  even. 

Def.  7.  An  odd  number  can  not  be  divided  by  2  with- 
out a  remainder.     Thus,  1,  3,  5,  7,  &c.,  are  odd. 

Rem. — All  even  numbers  except  2,  are  composite  numbers,  while 
odd  numbers  are  partly  prime  and  partly  composite. 

Def.  8.  A  divisor  of  a  number  will  exactly  divide  it; 
that  is,  without  a  remainder :  thus,  2  is  a  divisor  of  4 ; 
5  of  10,  &c. 

A  divisor  of  a  number  is  a  measure  of  that  number. 

Def.  9.  One  number  is  divisible  by  another,  when  the 
former  contains  the  latter  without  a  remainder.  Thus,  G  is 
divisible  by  2. 

Review. — 110.  "What  is  an  integer?  IIow  are  the  whole  numbers 
divided?  What  is  a  prime  number?  Give  examples.  What  a  com- 
posite? Give  examples.  When  are  two  numbers  prime  to  each  other? 
Give  examples.     What  is  an  even  number?     An  odd  ?     Give  examples. 


FACTORING.  121 

Def.  10.  A  multiple  (dividend),  of  a  number  is  the 
product  arising  from  taking  it  a  certain  number  of  times  : 
thus,  6  is  a  multiple  of  2,  because  it  is  equal  to  2  taken 
3  times.     Hence, 

A  multiple  of  a  number  can  be  divided  by  it  without  a  re- 
mainder.    Therefore,  every  multiple  is  a  composite  number. 

b 

Def.  11.  A  factor  of  a  number  is  a  number  that  will 

exactly  divide  it :  thus,  4  is  a  factor  of  8,  12,  16,  &c. 

Rem. — The  terms,  factor,  divisor  and  measure,  all  mean  the  same 
thing.  Every  composite  number  being  the  product  of  two  or  more 
factors,  each  factor  must  exactly  divide  it,  (Art.  37). 

Hence,  every  factor  of  a  number,  is  a  divisor  of  that  number. 

Def,  12.  A  pra?ie  factor  of  a  number  is  a  prime  number 
that  will  exactly  divide  it :  thus,  3  is  a  prime  factor  of  12  ; 
while  4  is  a  factor  of  12,  but  not  a,  prime  factor. 

Therefore,  all  the  prime  factors  of  a  number,  are  all  the  prime 
numbers  that  will  exactly  divide  it:  thus,  I,  3,  and  5,  are  all  the 
prime  factors  of  15. 

Every  composite  number  is  equal  to  the  product  of  all  its 
prime  factors :  thus,  all  the  prime  factors  of  10  are  1,  2,  and  5; 
1X2X5  =  10. 

Def.  13.  An  aliquot  part  of  a  number,  is  a  number 
that  will  exactly  divide  it :  thus,  1,2,  3,  4,  and  6,  are 
aliquot  parts  of  12. 

RESOLVING  NUMBERS  INTO  PRIME  FACTORS. 

Art.  111.  The  smaller  composite  numbers  may  be 
resolved  into  their  prime  factors  by  inspection  ;  thus, 

6=2X3;  8=2X2X2;  9=3X3;  10=2X5. 

In  the  case  of  large  numbers,  their  factors  are  found  by 
trial ;    that  is,   by  dividing  by  each  of  the  prime   num- 


Review, — 110.  Rem.  Are  the  even  numbers  prime  or  composite  ?  Are 
the  odd  numbers  ?  What  is  a  divisor  of  a  number  ?  Give  examples.  When 
is  one  number  divisible  by  another  ?    Give  examples. 

110.  What  is  a  multiple  ?  Give  examples.  A  factor  ?  Give  examples. 
Rem.  What  terms  besides  divisor  are  used  in  the  same  sense  ?  Why  is 
every  factor  a  divisor  ?     What  is  a  prime  factor  ?     Give  an  example. 


122  RAY'S  PRACTICAL   ARITHMETIC. 

bers  2,  3,  5,  7,  &c. ;     the  prime  factors  of  any  number, 
being  all  the  prime  numbers  that  will  exactly  divide  it. 

In  determining  either  the  factors,  or  the  prime  factors, 
of  a  number,  observe  the  following  principles. 

Principle  1. — A  factor  of  a  number  is  a  factor  of  any 
multiple  of  that  number. 

Thus,  3  is  a  factor  of  6,  and  of  any  number  of  times  6; 
for,  6  is  2  threes,  and  any  number  of  times  6  will  be  twice  as 
many  times  3. 

Principle  2.  A  factor  of  any  two  numbers  is  also  a 
factor  of  their  sum. 

Since  each  number  contains  the  factor  a  certain  number  of 
times,  their  sum  must  contain  it  as  many  times  as  both  numbers. 

Thus,  2  being  a  factor  of  6  and  8,  it  is  a  factor  of  their  sum; 
for,  G  is  3  tioos,  and  8  is  4  twos,  and  their  sum  is  3  /icos -f-4 
twos,=^l  twos. 

Art.  112.  From  these  two  Principles,  are  derived 
SIX    PROPOSITIONS. 

Prop.  1.  Every  number  ending  with  0,  2,  4,  6,  or  8, 
is  divisible  by  2. 

Illustration. — Every  number  ending  with  0,  is  either  10  or 
some  number  of  tens;  and,  since  10  is  divisible  by  2,  any  num- 
ber of  tens  will  be  divisible  by  2.     Prin.  1. 

Again:  any  number  ending  with  2,  4,  6,  or  8,  maybe  considered 
a  certain  number  of  tens,  plus  the  figure  in  units'  place : 

And,  as  each  of  the  two  parts  of  the  number  is  divisible 
by  2,  therefore,  Prin.  2,  the  number  itself  is  divisible  by  X 

Conversely :  No  number  is  divisible  by  2,  unless  it  ends  with 
a  0,  2,  4,  0,  or  8. 

Prop.  II.  Every  number  is  divisible  by  4,  when  the 
number  denoted  by  its  first  two  digits  is  divisible  by  4. 

Review. — 110.  What  ia  an  aliquot  part  of  a  number?  Give  examples. 
111.  How  may  the  smaller  composite  numbers  be  resolved  into  prime  fac- 
tors ?     What  are  the  prime  factors  of  G  ?    Of  8  ?     Of  9  ?     OflO? 

111.  In  determining  the  factors  of  a  number,  what  two  principles  are 
used  ?    Explain  the  first  principle.     The  second. 


FACTORING.  123 

Illustration. — Since  100  is  divisible  by  4,  any  number  of  hun- 
dreds is  divisible  by  4 ;  and  any  number  of  more  than  two  places 
of  figures,  may  be  regarded  as  a  certain  number  of  hundreds,  plus 
the  number  denoted  by  the  first  two  digits. 

Then,  since  both  parts  of  the  number  are  divisible  by  4,  Prin.  2, 
the  number  itself  is  divisible  by  4. 

Conversely :  No  number  is  divisible  by  4,  unless  the  number 

denoted  by  its  first  two  digits  is  divisible  by  4. 

t 

Prop.  III.  Every  number  is  divisible  by  5,  when  its 
right  hand  digit  is  0  or  5. 

Illustration. — Ten  being  divisible  by  5,  and  every  number 
consisting  of  two  or  more  places  of  figures,  being  composed  of  tens, 
plus  the  figure  in  the  units'  place: 

Therefore,  if  this  is  5,  both  parts  of  the  number  are  divisible 
by  5 ;  hence,  Prin.  2,  the  number  itself  is  divisible  by  5. 

Conversely :  No  number  is  divisible  by  5,  unless  its  right  hand 
digit  is  0  or  5. 

Prop.  IY.  Every  number  whose  first  digits  are  0,  00, 
&c.,  is  divisible  by  10,  100,  &c. 

Illustration. — If  the  first  figure  is  a  cipher,  the  number  is 
either  10,  or  some  multiple  of  10;  and, 

If  the  first  two  figures  are  ciphers,  the  number  is  either  100,  or 
some  multiple  of  100 ;  hence,  Prin.  1,  the  proposition  is  true. 

Conversely:  No  number  is  divisible  by  10,  100,  &c.,  unless 
it  ends  with  0,  00,  &c. 

Prop.  V.  Every  composite  number  is  divisible  by  the 
product  of  any  two  or  more  of  its  prime  factors. 

Illustration.— Thus,  the  number  30  is  equal  to  2x3X5;  now, 
if  30  be  divided  by  the  product  of  either  two  of  the  factors,  the 
quotient  must  be  the  other  factor;  if  no^  so,  the  product  of  the 
three  factors  would  not  be  30 :  and, 

The  same  may  be  shown  of  any  other  composite  number. 


Review.— 112.  When  is  a  number  divisible  by  2  ?  Why  ?  When  not 
divisible  by  2  ?  When  is  a  number  divisible  by  4  ?  Why  ?  When  not 
divisible  by  4?  When  is  a  number  divisible  by  5  ?  Why  ?  When  not 
d-'visible  by  5  ?  When  is  a  number  divisible  by  10,  100,  &c.  ?  Why? 
When  not  divisible  by  10,  100,  tfec.  ? 


124  RAY'S    PRACTICAL    ARITHMETIC. 

It  follows,  from  Prop.  5,  that  if  any  even  number  is  divisible 
by  3,  it  is  also  divisible  by  6.  For,  if  an  even  number,  it  is 
divisible  by  2 ;  and,  being  divisible  by  2  and  by  3,  it  is  also 
divisible  by  their  product,  2X3,  or  6. 

Prop.  VI.  Every  prime  number,  except  2  and  5,  ends 
with  1,  3,  7,  or  9 :  a  consequence  of  Prop.  1  and  3. 

Art.  113.  1.  What  are  the  prime  factors  of  30? 

Solution. — If  2  is  exactly  contained  in  30,  it     operation. 
will  be  a  factor  of  30.    By  trial,  it  is  found  to  be  a  2)30 

factor.     Again,  Q  vT^ 

If  3  exactly  divides  30,  it  will  be  a  factor  of  it ;  ^)  i-^ 

but,  since  30  is  2  times  15,  if  3  is  a  factor  of  15,  it  5 

will  also  be  a  factor  of  30,  (Art.  Ill,  Prin.  1.) 

To  ascertain  if  3  is  a  factor  of  30,  see  if  it  is  a  factor  of  15.  Trial 
shows  that  3  is  a  factor  of  15  ;  hence,  it  is  a  factor  of  30. 

For  the  same  reason,  whatever  number  is  a  factor  of  5,  is  a  fac- 
tor of  15  and  30;  but  5  is  a  prime  number,  having  no  factor  except 
itself  and  unity  ;  hence,  the  prime  factors  of  30,  are  1,  2,  3,  and  5. 

2.  Find  the  prime  factors  of  42.  Ans.  1,  2,  3,  7. 

3.  Find  the  prime  factors  of  70.  Ans.  1,  2.  5,  7. 

RULE 

FOR    RESOLVING   A    COMPOSITE   NUMBER   INTO  PRIME   FACTORS- 

Divide  the  given  number  by  any  pnme  number  that  will 
exactly  divide  it;  divide  the  quotient  in  the  same  manner^  and 
so  continue  to  divide,  until  a  quotient  is  obtained  which  is  a 
prime  number ;  the  last  quotient  and  the  several  divisors  will 
constitute  the  prime  factors  of  the  given  number. 

Rem. — 1.  It  will  generally  be  most  convenient  to  divide,  first 
by  the  smallest  prime  number  that  is  a  factor. 


Review. — 112.  By  what  is  every  composite  number  divisible  ?  Why  ? 
When  any  even  number  is  divisible  by  3,  by  what  is  it  also  divisible? 
With  what  figures  do  all  prime   numbers,  except  2  and  5,  terminate  ? 

113.  Find  the  prime  factors  of  80,  and  explain  the  process.  What  is 
the  rule  for  resolving  a  number  into  prime  fivctors? 


FACTORING. 


125 


2.  The  least  divisor  of  any  number  is  a  prime  number;  for,  if  it 
were  composite,  it  might  be  separated  into  factors,  which  would  be 
still  smaller  divisors  of  the  given  numbers.   Art.  Ill,  Prin.  1, 

Hence,  the  prime  factors  of  any  number  may  be  found,  by  first 
dividing  it  by  the  least  number  that  will  exactly  divide  it ;  then 
divide  the  quotient  as  before,  and  so  on. 

3.  Since  1  is  a  factor  of  every  number,  either  prime  or  composite, 
it  is  not  usually  specified  in  reckoning  the  factors  of  a  number. 


SEPARATE    INTO 


4. 

8. 

Ans.  2,  2,  2. 

13. 

5. 

12. 

Ans.  2,  2,  3. 

14. 

6. 

14. 

Ans.  2,  7. 

15. 

7. 

15. 

Ans.  3,  5. 

16. 

8. 

16. 

Ans.  2,  2,  2,  2. 

17. 

9. 

18. 

Ans.  2,  3,  3. 

18. 

10. 

20. 

Ans.  2,  2,  5. 

19. 

11. 

22. 

Ans.  2,  11. 

20. 

12. 

24. 

Ans.  2,  2,  2,  3. 

21. 

PRIME   FACTORS, 

26.  Ans.  2,  13. 

27.  Ans.  3,  3,  3. 

28.  Ans.  2,  2,  7. 
30.  Ans.  2,  3,  5. 
32.  Ans.  2,  2,  2,  2,  2. 

34.  Ans  2,  17. 

35.  Ans.  5,  7. 
66.  Ans.  2,  3,  11. 


98. 


Ans.  2,  7,  7. 


22.  "What  are  the  prime  factors  of  105?       Ans.  3,  5,  7. 

23.  Of  168?        Ans.  2,2,  2,3,  7. 

24.  216?        ^«s.  2,  2,  2,  3,  3,  3. 

25.  330?        ^ws.  2,  3,  5,  11. 

To  find  the  prime  factors  common  to  two  numbers,  resolve  each 
into  prime  factors:  then  take  the  factors  common  to  both. 


WHAT   PRIME    FACTORS   ARE   COMMON 

26.  To  110  and  210? Ans.  2,  5. 

27.  105  and  231? Ans.  3,7. 

28.  330  and  390  ? Ans.  2,  3,  5. 

29.  231  and  330? .472s.  3,  11. 

Eeview. — 113.  Eem,  1.  What  prime  factor  should  be  first  taken  as  a 
divisor  ?     2,  Why  is  the  least  divisor  of  any  number  a  prime  number  ? 

Eem.  3.  Why  is  unity  not  reckoned  among  the  prime  factors  of  a  number  ? 
How  may  the  prime  factors  common  to  two  numbers  be  found  ? 


126 


RAY'S    PRACTICAL   ARITHMETIC. 


Art.  114.  Since  any  composite  number  is  divisible  not 
only  by  each  of  its  prime  factors,  but  also  by  the  product 
of  any  two  or  more  of  them,  (Art.  112,  Prop.  V.), 

Therefore,  to  find  the  several  divisors  of  a  composite 
number,  resolve  it  into  its  prime  factors,  and  form  from 
them  as  many  different  products  as  possible. 

Thus,  30  =  2  X  3  X  5,  and  its  several  divisors  are  2,  3,  5, 
and  2X3,  2X5,  and  3X5. 

WHAT   ARE   THE   SEVERAL   DIVISORS 

1.  Of    42?        Ans.  2,  3,  7,6,  14,21. 

2.  105?        J7?s.  3,  5,  7,  15,  21,  35. 

3.  20?        Ans.  2,  5,  4,  10. 

4.  24?        .4«.s.  2,  3,  4,  6,  8.  12. 

S^°  For  additional  problems,  see  Rays  Test  Examples. 


IX.   GREATEST    COMMON    DIVISOR. 

Art.  115.  A  divisor  or  measure  of  a  number  (Art.  110, 
Def.  8),  is  a  number  that  will  divide  it  without  a  re- 
mainder ;    thus,  2  is  a  divisor  of  4 ;     3  of  6,  &c. 

A  common  divisor  of  two  or  more  numbers,  is  a  num- 
ber that  will  divide  each  without  a  remainder ;  thus,  2  is 
a  common  divisor  of  12  and  18. 

The  greatest  common  divisor  of  two  or  more  numbers, 
is  the  greatest  number  that  will  divide  each  without 
a  remainder ;  thus,  6  is  the  greatest  common  divisor 
of  12  and  18. 

Rem. — Two  numbers  may  have  several  common  divisors,  but 
only  one  greatest  common  divisor. 

G.  C.  D.  should  be  read,  greatest  common  divisor. 

Review. — 114.  How  may  tho  several  divisors  of  a  composito  number 
bo  found?     Why?     115.  What  is  a  divisor  of  a  number  ?     Give  exiimples. 

1  If).  What  is  a  c(/mm,on  divisor  of  two  or  more  numbers  ?  Give  example. 
What  tho  (jreiUest  common  divisor  ?     Give  example. 


GREATEST   COMMON    DIVISOR.  127 

Art.  116.  To  find  the  greatest  common  divisor  of  two 
numbers. 

FIRST   METHOD. 

1.  Find  the  greatest  common  divisor  of  70  and  154. 

Solution, — Resolving  the  numbers  into 
their  prime  factors,  by   inspection  or  by  ^^   ^^^  ^' 

the   rule    (Art.    113),    shows    that    70=         70  =  2x5X     7 
2X5X7,  and  154  =  2X7X11.  154  =  2X7x11 

Since  2   and  7  are  factors  of  each  of  Ans    9  y  7       14 

the  numbers,  both  may  be  exactly  divided 
by  2  or  7,  or  by  their  product :  2X7=14. 

As  2  and  7  are  the  only  prime  factors  common  io  the  numbers, 
no  number  except  2,  7,  and  their  product,  2  X  7  =  14,  will  exactly 
divide  both  of  them  :   therefore,  2  X  7  =  14,  is  the  G.  C.  D. 

2.  Find  the  G.  C.  D.  of  6  and  10.      ...     Ans.  2. 

3.  Of  30  and  42 Ans.  2X3  =  6. 

Jk  Rule  I. — Resolve  the  given  niimhers  into  prime  factors ;  the 
product  of  the  factors  which  are  common,  will  be  the  greatest 
common  divisor. 

Rem. — The  greatest  com.  divisor  of  two  numbers  contains,  as 
factors,  all  the  other  com.  divisors  of  those  numbers.  Thus,  6,  the 
greatest  com.  divisor  of  30  and  42,  contains,  as  factors,  2  and  3,  the 
only  remaining  com.  divisors  of  those  numbers. 

WHAT   IS   THE   GREATEST   COMMON   DIVISOR 

4.  Of  42  and    54? ^ns.  2X3=    6. 

5.  70  and  110? ^ws.  2x5  =  10. 

6.  105  and  165? ^ns.  3X5  =  15. 

7.  60  and    90  ?      .     .     .     .  Ans.  2  X  3  X  5  =  30. 
1^      8.       140  and  210?      .     .     .     .  ^tis.  2  X  5  X  7  =  70. 

9.         66  and  154? ^/?s.  2  X  11  =22. 

10.  154and2S0? Ans.  2X7  =  U. 

11.  231  and  273? ^ns.  3X7  =  21. 

Review, — 115,  Rem,  Can  two  numbers  have  more  than  one  common 
divisor?  116.  Find  the  G.  C,  D,  of  SO  and  42,  by  separating  each  into 
prime  factors.      "What  is  Rule  I  ? 

116.  Rem.  "What  factors  does  the  G.  C.  D.  of  two  numbers  contain  ? 


128  RAY'S   PRACTICAL   ARITHMETIC. 

Note. — When  there  are  more  than  two  numbers,  resolve  each 
into  prime  factors;  then  take  the  product  of  the  common  factors. 

WHAT   IS   THE   GREATEST   COMMON   DIVISOR   OP 

12.  30,  42,  and  66?        .     .     .     ^«s.  2X3=6. 

13.  60,  90,  and  150?      .     .  ^tis.  2  X  3  X  5  =  30. 
When  the  numbers  are  large,  it  is  better  to  adopt 

Art.  117.    THE   SECOND   METHOD. 
This  method  depends  on  the  following  principles: 

1st  Prin.  a  divisor  of  a  numher^  is  a  divisor  of  any 
multiple  of  that  number;  Art.  Ill,  Prin.  1. 

2d  Prin.  A  common  divisor  of  two  numbers,  is  a  di- 
visor of  their  SUM ;  Art.  Ill,  Prin.  2. 

3d  Prin.  A  common  divisor  of  two  numbers,  is  a  divisor 

of  their  DIFFERENCE. 

Since  each  of  the  numbers  contains  the  com.  divisor  a  certain 
number  of  times,  their  difference  must  contain  it  as  many  times 
as  the  larger  contains  it  more  times  than  the  smaller. 

Thus,  2,  being  a  divisor  of  14  and  8,  must  be  a  divisor  of 
their  difference;  for,  14  is  7  twos,  and  8  is  4  ticos,  and  their 
difference  is  7  twos  minus  4  twos=^S  twos. 

Therefore,  if  a  number  be  separated  into  two  parts,  any 
number  which  will  exactly  divide  the  given  number  and  one  of 
its  parts,  will  also  exactly  divide  the  other. 

In  this  case,  either  of  the  parts  is  the  difference  between  the 
given  number  and  the  other  part. 

4th  Prin.  The  greatest  common  divisor  of  tico  numbers, 
is  a  divisor  of  their  remainder  after  division.     See  Solu. 

1.  Find  the  G.  C.  D.  of  16  and  44. 

Solution. — As   16  is  a  divisor   of  operation. 

itself,  if  it  be  a  divisor  of  44,  it  will         16)44(2 
be  the  G.   C.   D.   required,    since   no  ^-' 


number   can    have  a    divisor   greater  1  2 ")  1  6  T  1 

than    itself.       But    16    is    contained  j  2 

twice   in   44,  with    a  remainder  12;  

hence,  16  is  not  the  G.  C.  D.     Since  4)12(3 

the  G.  C.  D.  is  a  divisor  of  16  and  44,  by  first  Prin.,  it  will  be  a 


I 


GREATEST    COMMON  DIVISOR.  129 

divisor  of  16X2  =  32;  hence,  by  3d  Prin.,  it  must  be  a  divisor 
of  44  —  32  =  12;  that  is,  the  G.  C.  D.  of  two  nuvibers  is  also  a 
divisor  of  their  remainder  after  division. 

Hence,  the  G.  C.  D.  of  16  and  44  is  also  a  com.  divisor  of  12 
and  16,  and  it  can  not  exceed  12,  Since  12  is  a  divisor  of  itself, 
if  it  be  a  divisor  of  16,  it  must  be  the  G.  C.  D.  sought, 

By  dividing  12  into  16,  the  remainder  is  4 ;  hence,  12  is  not  the 
G.  C.  D.;  but,  by  Prin.  4th,  the  G.  C.  D.  of  12  and  16  is  a  divisor 
of  4,  their  remainder  after  division;  hence,  the  G.  C.  D,  of  16  and 
44  can  not  exceed  4,  and  must  be  a  divisor  of  4  and  12, 

By  dividing  12  by  4,  there  is  no  remainder;  hence,  4  is  a  divisor 
of  12,  and  therefore,  of  12  X  1-|-4  =  16,  Prin,  2d.     And, 

Since  4  is  a  divisor  of  12  and  16,  it  must  be  a  divisor  of 
16  X2-|- 12  =  44;  and  since  the  G.  C,  D,  can  not  exceed  4,  aud  4 
is  a  divisor  of  16  and  44,  therefore,  4  is  the  G.  C,  D.  souglil, 

2.  What  is  the  G.  C.  D.  of  14  and  35?  Ans.  7. 

3.  What  is  the  G.  C.  D.  of    9  and  24?  Ans.  3. 

Rule  II. — Divide  the  greater  number  hij  the  less,  and  that 
divisor  by  the  remainder,  and  so  on,  alicays  dividing  the  last 
divisor  by  the  last  remainder,  till  nothing  remains;  the  last 
divisor  will  be  the  greatest  common  divisor. 

Note, — To  find  the  G,  C,  D.  of  more  than  two  numbers,  first  find 
the  G,  C.  D.  of  two  of  them,  then  of  that  com,  divisor  and  one  of 
the  remaining  numbers,  and  so  on  for  all  the  numbers;  the  last 
com.  divisor  will  be  the  G.  C,  D.  of  all  the  numbers, 

FIND   THE   GREATEST   COMMON   DIVISOR    OP 

8.  337  and  759.  Ans.   1. 
-9.  9873  and  69087.  ^tjs.  3. 

10.  1814  and  1776,  Ans.  2. 

11.  693  and  1815.  ^«s.  33. 

Eetikw. — 117.  On  what  principles  does  the  second  method  of  finding 
the  G.  C.  D.  depend  ?  Explain  the  third  principle.  Find  the  G.  C.  D. 
of  9  and  24,  and  explain  the  fourth  principle.     What  is  Eule  II  ? 

117.  Note.  How  is  the  G,  C,  D.  of  more  than  two  numbers  found  ? 
3d  Bk.  9 


4. 

42  and  495, 

Ans. 

3. 

5, 

247  and  323, 

Ans. 

19. 

6. 

285  and  465. 

Alls. 

15, 

7. 

532  and  1274. 

Ans. 

14. 

130 


RAY'S    PRACTICAL    ARITHMETIC. 


12.  Find  the  G.  C.  D.  of  2145  and  3171, 

13.  Of  810  and  17017. 
11.       Of        66281  and  153152. 

15.  Of      40,  55,  and  105.    . 

16.  Of    70,  154,  and  819.    . 

17.  Of  120,  168,  and  1768. 


.  Ans.  39. 

.  Alls.  7. 
Ans.  908. 
.  Alts.  5. 
.  Alls.  7. 
.  Ans.    8. 


For  additional  problems,  sec  Ray's  Test  Examples. 


X.    LEAST  COMMON  MULTIPLE. 

Art.  118.  A  midtiple  (dividend),  of  a  number,  is  a 
number  that  can  be  divided  by  it  without  a  remainder. 

Thus,  12  is  a  multiple  of  3,  because  3  is  contained  in  12 
an  exact  number  of  times,  4.     Art.  110,  Def.  10. 

A  common  multiple  (dividend),  of  two  or  more  num- 
bers, is  a  number  that  can  be  divided  by  each^  without  a 
remainder. 

Thus,  24  is  a  common  multiple  of  3  and  4. 

The  hast  common  multiple  of  two  or  more  numbers,  is 
the  least  number  that  can  be  divided  by  each  without  a 
remainder. 

Thus,  12  is  the  least  common  multiple  of  3  and  4. 

Rem. — 1.  As  the  Com.  Mul.  of  two  or  more  numbers  contains 
each  of  them  as  a  factor,  it  is  a  composite  number. 

2.  As  the  continued  product  of  two  or  more  numbers  is  divisible 
by  each  of  them,  a  Com.  Mul.  of  two  or  more  given  numbers  may 
always  be  found  by  taking  their  continued  product;    and, 

Since  any  multiple  of  this  product  will  be  divisible  by  each  of 
the  given  numbers,  (Art.  Ill,  1st  Prin.),  an  unlimited  number  of 
Com.  Multiples  may  be  found  for  any  given  numbers. 

Review. — 118.  What  is  a  multiple  of  a  number?  Give  an  example. 
What  is  a  armmon  multiple  of  two  or  more  numbers?  What  the  least 
common  multiple  ?  Rem.  1.  Is  a  common  multiple  of  two  or  more  numbers, 
a  prime  or  composite  number?  Why  ?  2.  How  may  a  Com.  Mul.  always 
bo  found  ?     How  many  Com.  Multiples  may  numbers  have  ?     Why  ? 


LEAST    COMMON    MULTIPLE.  131 

Art.  119.  To  find  the  least  common  multiple  of  two  or 
more  numbers. 

FIRST    METHOD. 

One  number  is  divisible  by  another,  when  it  contains 
all  the  prime  factors  of  that  number. 

Thus,  30  is  divisible  by  0,  because  30^=2X3X5,  and 
6  =  2X3;  the  prime  factors  of  6,  which  are  2  and  3,  being 
also  factors  of  30. 

One  number  is  not  divisible  by  another,  unless  it  con- 
tains all  the  prime  factors  of  that  other. 

Thus,  10  is  not  divisible  by  6,  because  3,  one  of  the  prime 
factors  of  6,  is  not  a  factor  of  10. 

Hence,  a  common  multiple  of  two  or  more  numbers  must 
contain  all  the  prime  factors  in  those  numbers;    and. 

To  be  the  least  common  multiple,  (L.  C.  M.),  it  must  not  con- 
tain any  prime  factor  not  found  in  some  one  of  the  numbers. 

B@^L.  C.  M.  should  be  read,  least  common  multiple. 

1.  What  is  the  L.  C.  M.  of  6  and  10? 
Solution. — By  factoring,  6  =  2X3, 

1    in        .)  NX-         A  u  a  OPERATION. 

and  10  =  2  X  o.     A  number  composed  n 9  v  '-^ 

of  the  factors  2,  3,  and  5,  will  contain  -,  r. . .  ^  p. 

all  the  factors  in  each  of  the  numbers 

6   and   10,  and  will    contain  no  other     2X3X5^=30  Ans. 

factor;    therefore,  cross  out  (cancel)  the 

factor  2,  in  one  of  the  numbers;    the    product  of  the  remaining 

factors,  2X3  X  5  =  80,  will  be  the  L.  C.  M.  of  6  and  10. 

2.  What  is  the  L.  C.  M.  of  6,  8,  and  12? 

Solution. — By    factoring     the 
V         *v,         •        ^    .      o                           operation. 
numbers,  the  prime   factor  2  oc-  n 0  v  q 

curs  once  in  6,  three  times   in  8,  q 9  v  9  v  9 

and  twice  in  12 ;    hence   it  must  12 0v0v3i 

occur  three  times,  and  only  three 

times,  in  the  L.  C.  M. ;    therefore,      3X2X2X2  =  24  Ans. 

after   reserving    it    as    a    factor 

three  times,  cancel  it  in  the  other  numbers. 

The  prime  factor  3  occurs  once  in  6  and  once  in  12;    hence,  it 

must  occur  once,  and  only  once,  in  the  L.  C.  M.     After  reserving 

it  once  as  a  factor,  cancel  the  other  factor  3.     The  L.  C.  M.  is 

then  found  by  multiplying  together  the  figures  not  canceled. 


5. 

6,    8,    9. 

Ans.    72 

6. 

6,  15,  35. 

Ans.  210 

7. 

10,  12,  15. 

Ans.    60 

132  RAY'S  PRACTICAL   ARITHMETIC. 

3.  Find  the  L.  C.  M.  of  12  and  30.      .     .  Ans.  60. 

4.  Of  6,  10,  and  18 Ans.  90. 

Kule  I. — Separate  the  numbers  into  prime  factors ;  then 
multiply  together  only  such  of  those  factors^  as  are  necessary  to 
form  a  product  that  will  contain  all  the  prime  factors  in  each 
number^  using  no  factor  oftener  than  it  occurs  in  any  one 
number. 

Note. — The  solution  of  Ex.  2,  shows  that  the  same  factor  must 
be  taken  the  greatest  number  of  times  it  occurs  in  either  number. 
After  factoring,  cancel  (cross  out)  the  needless  factors. 

FIND   THE    LEAST   COMMON   MULTIPLE   OP 

8.  9,  15,  18,  24.  Ans.  360. 

9.  8,  15,  12,  30.  Ans.  120. 
10.  14,  21,  30,  35.  Ans.  210. 

SECOND   METHOD. 

Art.  120.  The  L.  C.  M.  of  two  or  more  numbers, 
contains  all  the  prime  factors  of  each  of  the  numbers 
07ice,  and  no  other  factors. 

For,  if  it  did  not  contain  all  the  prime  factors  of  any  num- 
ber, it  would  not  be  divisible  by  that  number;  and,  if  it  con- 
tained any  prime  factor  not  found  in  either  of  the  numbers, 
it  would  not  be  the  least  common  multiple. 

Thus,  the  L.  C.  M.  of  4  (2X2),  and  G  (2X3),  must  con- 
tain the  factors  2,  2,  3,  and  no  others. 

1.  Find  the  L.  C.  M.  of  6,  9,  and  12. 

Solution. — Arranging  the  num-                       operation. 
bers   as  in   the  margin,    we   tind                    2)6      9      12 
that  2  is  a  prime  factor  common                    qV^      Q        T 
to  two  of  them.  1 

Hence,  2  must  be    a   factor  of  13        2 

the  L.  C.  M.;  therefore,  place  it  9  ^  o  ^  -^  v  9  _  Qfi  J  .c 
on  the  left,  and  cancel  it  in  the  -A^AciA-  —  i)0  .i«5. 
numbers  of  which  it  is  a  factor,  by  dividing  by  it. 

Next,  observe  that  3  is  a  factor  common  to  the  quotients  and 
the  remaining  number,  and  hence,  (Art.  Ill,)  is  a  factor  of 
the  given    numbers,    and   must    be    a    factor  of   the   L.    C.    M 


LEAST   COMMON   MULTIPLE.  133 

therefore,  place  it  oa  the  left,  and  cancel  it  in  each  of  the  numbers 
in  the  2d  line,  by  dividing  by  it.  As  the  numbers  3  and  2,  in  the 
3d  line,  have  no  common  factors  to  cancel,  we  do  not  divide  them. 

Thus  we  find,  that  2,  3,  3,  and  2,  are  all  the  prime  factors  in 
the  given  numbers;  hence,  their  product,  2X3X3X2  =  36  is 
the  L.  C.  M.  of  6,  9,  and  12. 

In  this  operation,  let  the  learner  notice, 

1st.  The  number  36  is  a  common  multiple,  because  it  con- 
tains all  the  prime  factors  in  each  of  the  numbers ;  it  is  the  least 
C.  M.,  because  all  the  needless  factors  were  canceled  by  dividing. 

2d.  To  cancel  needless  factors,  divide  by  a^rme  number. 

By  dividing  by  a  composite  number,  in  some  cases,  all  the 
needless  factors  are  not  canceled ;  thus,  in  the  preceding  ex- 
ample, 6  will  exactly  divide  two  of  the  numbers  ;   but. 

In  dividing  by  6,  a  factor,  3,  is  left  uncanceled  in  the  mul- 
tiple 9,  and  thus  the  L.  C.  M.  is  not  obtained. 

2.  Find  the  L.  C.  M.  of  6  and  10.    .     .     .     Ans.  30. 

3.  Of  15,  21,  35 Ans.  105. 

Rule  II. — 1.  Place  the  mimhers  in  a  line,  divide  by  any 
prime  number  that  will  divide  two  or  more  of  them  without 
a  remainder,  and  place  the  quotients  and  undivided  numbers 
in  a  line  beneath. 

2.  Divide  this  line  as  before  :  continue  to  divide  till  no  number 
greater  than  1  will  exactly  divide  two  or  more  of  the  numbers. 

3.  Multiply  together  the  divisors  and  the  numbers  in  the  lowest 
line,  and  their  product  will  be  the  least  common  multiple. 

Rem. — If  the  given  numbers  contain  no  common  factor,  their 
product  will  be  the  L.  C.  M.  Thus,  the  L.  C.  M.  of  4,  5,  and  9, 
is  4X5X9  =  180. 

Eeview. — 119.  When  is  one  number  divisible  by  another?  Give  an 
example.  When  not  divisible  ?  Give  an  example.  What  actors  must 
the  Com.  Mul.  contain  ? 

119.  What  prime  factors  must  the  L.  C.  M.  not  contain  ?  Find  the 
L.  C.  M.  of  6,  10,  and  18,  and  explain  the  operation.     What  is  Eule  I? 

119.  Note.  How  often  must  the  same  factor  be  found  in  the  L.  C.  M.  ? 


134  RAYS   PRACTICAL   ARITHMETIC. 


FIND  THE   LEAST   COMMON   MULTIPLE   OP 


4.  9,  12.  Ans.  36. 

5.  14,  21.  Ans.  42. 

6.  6,  9,  15.  Ans.  90. 

7.  4,  14,  35.  Am.  140. 


8.  6,  10,  15,  18.     Ans.  90. 

9.  7,  11,  13,  3.  ^«s.  3003. 

10.  63,  12,  84,  7.     Ans.  252. 

11.  54,  81,  63.       Ans.  1134. 


12.  Of  8,  12,  20,  24,  25 Ans.      600. 

13.  9,  10,  24,  25,  32,  45.      ...  Ans.    7200. 

14.  98,  72,  64,  21,  18 Ans.  28224. 

15.  2,  3,  4,  5,  6,  7,  8,  9.     .     .     .  Ans.    2520. 


XI.   COMMON   FRACTIONS. 

Art.  121.  A  single  thing,  (Art.  1),  is  called  a  unit,  or 
owe,  which  may  be  divided  into  equal  parts. 

Thus,  suppose  3  apples  are  to  be  equally  divided  between 
2  boys:  after  giving  one  to  each,  there  would  remain  one  to 
be  divided  into  two  equal  parts,  to  complete  the  division. 

The  equal  parts  into  which  a  unit  is  divided  are  Jractions. 

Art.  122.  When  a  unit,  or  single  thing,  is  divided 
into  two  equal  parts,  one  of  the  parts  is  one-half. 

If  it  is  divided  into  three  equal  parts,  one  of  the  parts 
is  one-third;  two  of  the  parts,  two-thirds. 

If  divided  into  four  equal  parts,  one  of  the  parts  is 
one-fourth;  two  of  the  parts,  two-fourths;  and  three 
of  the  parts,  threefourths. 

If  divided  into  ^re  equal  parts,  the  parts  are  fifths;  if 
into  six  equal  parts,  sixths,  and  so  on.     Hence, 

When  a  unit  is  divided  into  equal  parts,  the  parts  are  named 
from  the  number  of  parts  into  which  the  unit  is  divided. 

Review.— 120.  Ex.  1.  Why  divide  by  2  ?  By  8  ?  Why  multiply 
together  the  numbers  2,  3,  3,  nnd  2  ?  Why  is  86  a  Com.  Mul.  of  6,  9, 
and  12  ?  Why  the  least  ?  To  cancel  needless  factors,  why  Twt  divide  by 
a  componte  number  ?     What  is  Rule  II  ? 

120.  Rem.  If  the  numbers  contain  no  ecmmon  factor,  how  is  their  L.  C.  M. 
found?  121.  How  do  you  divide  3  apples  equally  between  2  boys?  When 
a  unit  is  divided  into  equal  parts,  what  arc  the  parts  called  ? 


I 


¥ 


COMMON   FRACTIONS.  135 

Art.  123.  The  value  of  one  of  the  parts  depends  ou 
the  number  of  parts  into  which  the  unit  is  divided. 

Thus,  if  3  apples  of  equal  size  be  divided,  one  into  2,  another 
into  3,  and  another  into  4  equal  parts,  the  thirds  will  be  less 
than  the  halves,  the  fourths  less  than  the  thirds. 

Art.  124.  Fractions  are  divided  into  two  classes, 
Common  and  Decimal. 

Common  Fractions. are  expressed  by  two  numbers,  one 
above  the  other,  with  a  horizontal  line  between  them. 

Thus,  one-half  is  expressed  by  ^  ;    two-thirds  by  g. 

The  number  helow  the  line  is  the  denominator :  it  denominates, 
or  gives  name  to  the  fraction.  It  shows  the  number  of  parts 
into  which  the  unit  is  divided. 

The  number  above  the  line  is  the  numerator :  it  numbers  the 
parts,  showing  how  many  parts  are  taken. 

Thus,  in  the  fraction  |,  the  denominator,  5,  shows  that  the 
unit  is  divided  into  j^i'e  equal  parts,  and  the  numerator,  3,  shows 
that  the  fraction  contains  3  of  those  parts. 

The  numerator  and  denominator  together,  are  called  the  terms 
of  the  fraction.     Thus,  the  terms  of  |,  are  3  and  5. 

Art.  125.   ANOTHER  METHOD. 

[  In  the  definition  of  numerator  and  denominator,  refer- 
ence is  had  to  a  unit  only.  This  is  the  simplest  method  of 
considering  a  fraction  ;    but,  there  is  another  mode  : 

Illustration. — To  divide  2  apples  equally  among  3  boys, 
divide  each  apple  into  three  equal  parts,  making  6  parts  in  all; 
then  give  to  each  boy  2  of  the  parts,  expressed  by  §. 

Review. — 122,  When  a  unit  is  divided  into  two  equal  parts,  what  is 
one  part  called  ?  When  divided  into  three  equal  parts,  what  is  one  part 
called  ?  What  two  parts  ?  When  divided  into  four  equal  parts,  what  is 
one  part  called  ?  Two  parts  ?  Three  parts  ?  When  a  unit  is  divided  into 
equal  parts,  from  what  are  the  parts  named  ? 

123.  On  what  does  the  value  of  one  of  the  parts  depend  ?  Which  is 
greater,  1-half  or  1-third  ?     1-third  or  1-fourth  ?     1-fourth  or  1-fifth  ? 

124.  Into  what  two  classes  are  fractions  divided  ?  How  are  common 
fractions  expressed  ?  What  is  the  number  below  the  line  ?  Why  ?  What 
the  number  above  the  line  ?     Why  ?     What  are  the  terms  ? 


136  RAY'S   PRACTICAL   ARITHMETIC. 

In  selecting  one  boy's  share,  take  the  2  parts  from  one  apple, 
or  1  part  from  each  of  the  two  apples ;    hence, 

I  expresses  either  2  thirds  of  1  thing,  or  1  third  of  2  things. 

Also,  I  expresses  3  fifths   of  1  thing,  or  1  fifth  of  3  things. 

Therefore,  the  numerator  of  a  fraction  expresses  the  number 
of  units  to  be  divided ;  and  the  denominator  the  divisor,  or  what 
part  is  taken  from  each.     Hence, 

A  fraction  is  expressed  in  the  form  of  an  unexecuted  division, 

In  which,      The  dividend  is  the  numerator; 
The  DIVISOR     is  the  denominator  ; 
The  quotient  is  the  fraction  itself 

i  is  the  quotient  of  1  (numerator) -j- 3  (denom.); 
A  is  the  quotient  of  4  (numerator) -i-  5  (denom.); 
I  is  the  quotient  of  7  (numerator)  -r-  6  (denom.). 

Art.  126.  Since  fractions  arise  from  division,  one  of 
the  si^ns  of  division  (Art.  40,)  is  used  in  expressing 
them  ;  the  numerator  being  written  above,  and  the  de- 
ijominator  below,  a  horizontal  line. 

Thus,  three-fifths  is  written  |;    two-sixths  is  written  |. 

TO   READ   COMMON   FRACTIONS. 

Bead  the  number  of  parts  taken,  as  expressed  by  the  numerator; 
then  the  size  of  the  parts,  as  expressed  by  the  denominator. 

TO    WRITE    COMMON    FRACTIONS. 

Write  the  numerator,  place  a  horizontal  line  below  it,  under 
which  write  the  denominator. 

Kem. — In  reading,  |  means  two-thirds  of  one.  There  are  two  other 
methods,  (Art.  125) :  thus,  |  may  bo  read,  one-third  of  two,  or  two  dividtd 
hy  three  /  but  these  methods  arc  rarely  used. 

FRACTIONS   TO    BE    READ. 

fi      ••>      4      Ifl     11        13        5_        fc      14      IQ      11      _4  1_6       _8 

7»     B»     9'      IP     l'^'      \1i     2l'     2  9'     4  5'     «5'    ^fi'     iDO'    24  0'     l24  3" 

Review. — 125.  What  do  two-thirds  express  ?  What  does  the  numerator 
of  a  fraction  express?  The  denominatdr ?  In  what  form  is  a  fraction 
expressed?  What  is  the  dividend  ?  The  divisor  ?  The  quotient?  Give 
examples.  12G.  What  sign  is  used  in  fractions  ?  W^hy  ?  How  is  a  com- 
mon fraction  read  ?     Give  examples.     How  written  ?     Give  examples. 


COMMON  FRACTIONS.  137 

FRACTIONS   TO   BE   WRITTEN. 


Three-sevenihs  : 
Five-ninths  : 
iiix-ie7iths : 
8eY  en-elevenths 


One-twelfth  : 
Two-thirteenths : 
^sine-sixteeuths : 
Elexen-twentieths 


Yoixrieen-twenty-ninihs  : 
ThiYty-one-niuett/-ihirds  : 
Twenty  -  three  -  one  -  hund- 
red-and-Jourths. 


If  an  orange  be  cut  into  8  equal  parts,  what  fraction 
will  express  3  of  the  parts  ? 

If  3  apples  be  divided  equally  among  4  boys,  what 
part  of  an  apple  will  be  given  to  each  boy? 

What  part  of  1  apple,  is  a  third  part  of  2  apples  ? 
What  expresses  the  quotient  of  5,  divided  by  8  ? 

Art.  127.  A  whole  number  may  be  expressed  in  the 
form  of  a  fraction,  by  writing  1  below  it  for  a  denominator. 

Thus,  2  may  be  written  f  and  is  read  two-ones. 

3  may  be  written  |  and  is  read  three-ones. 

4  may  be  written  |-  and  is  read  four-ones. 

But,  2  ones  are  2;  3  ones,  3,  &c.;  hence,  the  value  of  the 
number  is  not  changed. 

Art.  128.  If  2  apples  be  divided,  each  into  four  equal 
parts,  there  will  be  8  parts  in  all.  Three  of  the  parts 
(fourths)  are  expressed  by  | ;  4  parts  by  | ;  5  parts  by  |. 

When  the  number  of  parts  taken  is  less  than  the  number 
into  which  the  unit  is  divided,  the  value  expressed  is  less  than 
one,  or  the  whole  thing; 

When  the  number  of  parts  taken  is  equal  to  the  number 
into  which  the  unit  is  divided,  the  value,  as  |,  is  equal  to  1 ; 

When  greater  than  the  number  into  which  the  unit  is 
divided,  the  value,  |,  is  greater  than  1.     Hence, 

Review. — 126.  Of  the  fractions  to  be  read,  which  expresses  parts  of  the 
largest  size  ?  Which  the  smallest  ?  Which  the  least  number  ?  W'hich 
the  greatest  ?    W^hich  the  same  ?     Which  parts  of  the  same  size? 

127.  How  may  a  whole  number  be  expressed  in  fractional  form  ?  Does 
this  change  its  value  ?  Why  not  ?  128.  When  is  the  value  of  a  fraction 
less  than  1  ?     When  equal  ?     When  greater  ?    Illustrate  by  examples. 


138  RAY'S    PRACTICAL    ARITHMETIC. 

1st.  When  the  numerator  is  less  than  the  denominator,  ike 
value  of  the  fraction  is  less  than  1. 

2d.  When  the  numerator  is  equal  to  the  denominator,  the 
value  of  the  fraction  is  equal  to  1. 

3d.  When  the  numerator  is  greater  than  the  denominator, 
the  value  of  the  fraction  is  greater  than  1. 

DEFINITIONS. 

Art.  129.  1.  A  Fraction  is  an  expression  of  one  or 
more  of  the  equal  parts  of  a  unit,  or  one  thing. 

2.  A  Proper  Fraction  has  a  numerator  less  than  the  de- 
nominator;   as,  i,  I,  and  |. 

3.  An  Improper  Fraction  has  a  numerator  equal  to,  or 
greater  than  the  denominator;    as,  |,  and  |. 

Rem. — A  proper  fraction  is  so  termed,  because  it  expresses  a 
value  less  than  one.  An  improper  fraction  is  not  properly  a  fraction 
of  a  unit,  the  value  expressed  being  equal  to,  or  greater  than  one. 

4.  A  Simple  Fraction  is  a  single  fraction,  either  proper  or 
improper;    as,  i,  |,  and  |. 

5.  A  Compound  Fraction  is  a  fraction  of  a  fraction,  or  severjil 
fractions  joined  by  of;       as,  A  of  i,         f  of  |  of  |. 

6.  A  Mixed  Number  is  a  fraction  joined  to  a  whole  number; 
as,  2-^,  3i,  and  5|. 

7   A  Complex  Fraction  has  a  fraction  in  one  or  in  both  of 

3^  1  j  21 

its  terms  ;  as,      —  »  —  ?  — >  —  • 

PARTS   OF   FRACTIONS. 

Art.  130.  If  a  line,    ^ i ' f, H 

as  A  Ji,  be  divided  into 

two  equal  parts,  one  of  the  parts,  as  A  C,  is  termed  one- 
half  (A) :  that  is,  one-half  of  1,  or  the  whole  thing. 

Review. — 129.  What  is  a  fraction?  A  proper  fraction?  An  im- 
proper fraction?  Eem.  Why  is  a  proper  fraction  so  termed?  An  im- 
proper fraction?  What  i.s  a  simple  fraction  ?  A  compound  fraction  ?  A 
mixed  number  ?  A  complex  fraction  ?  Give  examples  of  each.  130.  What 
is  the  half  of  one-half  ?     Why  ?    The  third  of  one-half  ?     Why  ? 


COMMON   FRACTIONS.  139 

If  a  half  be  divided  into  3  equal  parts,  as  in  the  figure,  one 
of  the  parts  is  one-third  of  one-half,  expressed  thus,  ^  of  ^ ;  and 
this  is  one-sixth  of  the  whole  line :  that  is,  i  of  ^  is  ^. 

In  the  same  manner,  it  may  be  shown,  that  o  o^  ^  ^^  4 '  ^^^^ 
the  ^  of  I  is  I ;    that  |  of  i  is  ^;    that  2  ^f  3  is  jg :  tfce. 

If  one  apple  be  divided  into  three  equal  parts,  and  each  of 
these  parts  into  3  other  equal  parts,  into  how  many  parts  will 
the  whole  be  divided  ? 

What  part  of  the  whole  will  one  piece  be  ?     What  is  i  of  ^  ? 

If  one  orange  be  divided  into  3  equal  parts,  and  each  part 
into  4  other  equal  parts,  into  how  many  equal  parts  will  the 
whole  be  divided  ? 

What  part  of  the  whole  will  1  piece  be  ?    What  is  J  of  i  ? 
What  is  1  of  1;     lofi?     iofi?     J  of  4?     |  of  i? 
Young  Pupils  may  omit  the  Illustrations  until  they  review  the  book. 

GENERAL  PRINCIPLES. 

Art.  131.  If  the  numerator  of  the  fraction  f  be 
multiplied  by  3,  without  changing  the  denominator,  the 
result  will  be  f . 

m.  2  X  3  _  6 

IhUS,  Jj  —rj' 

Illustration. — Each  of  the  fractions  ^  and  ^  having  the  same 
denominator,  expresses  parts  of  the  same  size  :  but,  as  the  numerator 
of  the  second  fraction  (^),  is  3  times  that  of  the  first,  (^),  it  expresses 
3  times  as  many  of  those  equal  parts  as  the  first,  and  is  3  times  as 
large.     The  same  may  be  shown  of  any  other  fraction.     Hence, 

Proposition  1. — If  the  numerator  be  multiplied  witliout 
changing  the  denominator^  the  value  of  the  fraction  will  he  mul- 
tiplied as  many  times  as  there  are  units  in  the  multiplier. 

Hence,  a  fraction  is  multiplied^  by  multiplying  its  numerator. 

Review. — 131.  What  is  the  efifeet  of  multiplying  the  numerator  of  a 
fraction,  without  changing  the  denominator,  Prop.  I  ? 


140  RAY'S    PRACTICAL    ARITHMETIC. 

Art.  132.  If  the  numerator  of  the  fraction  4  be 
divided  by  2,  without  changing  the  denominator,  the 
result  will  be  i. 

Illustration. — Each  of  the  fractions  ^  and  |  having  the  same 
denominator,  expresses  parts  of  the  same  size ;  but  the  numerator 
of  the  second  fraction  (|),  is  only  one-half  the  numerator  of  the 
first  (|);  therefore,  |  expresses  one- half  as  many  of  those  equal 
parts  as  the  first  (^),  and  is  one-half  the  value.     Hence, 

Prop.  II. — If  the  numerator  he  divided  without  changing 
the  denominator,  the  value  of  the  fraction  will  be  divided  as 
many  times  as  there  are  units  in  the  divisor. 

Hence,  a  fraction  is  divided  by  dividing  its  numerator. 

Art.  133.  If  the  denominator  of,  the  fraction  |  be 
multiplied  by  2,  without  changing  the  numerator,  the 
result  will  be  |. 

Thus,       4  X  2  "^  8  ' 

Illustration. — Each  of  the  fractions  |  and  |  having  the 
same  numerator,  denotes  the  same  number  of  parts;  but,  in  the 
second  (g)  the  parts  are  one-half  the  size  of  those  in  the  first  (|): 
but,  5=^2  ^^  4  (■^^*-  130);  consequently,  the  ivhole  value  of  the 
second  fraction  is  one-half  that  of  the  first.     Hence, 

Prop.  III.  If  the  denominator  be  multiplied  icithont  changing 
the  numerator,  the  value  of  the  fraction  will  be  divided  as  many 
times  as  there  are  units  in  the  multiplier. 

Hence,  a  fraction  is  divided,  by  multiplying  its  denominator. 

Art.  134.  If  the  denominator  of  the  fraction  §  be 
divided  by  3,  without  changing  the  numerator,  the  result 
will  be  i. 

2  ^ 

Thus,       9_3-3' 

Illustration. — Each  of  the  fractions  ^  and  |  having  the 
same  numerator,  denotes  the  same  number  of  parts;  but,  in  the 


I 


COMMON   FKACTIONS.  141 

second  fraction  (|),  the  parts  are  3  times  the  size  of  those  in  the 
lirst  (J):  but,  ^  =  ^  of  i  (Art.  130);  consequently,  the  value  of 
the  second  fraction  is  3  times  that  of  the  first.     Hence, 

Prop.  IV.  If  the  denominator  he  divided^  loithout  changing 
the  numerator,  the  value  of  the  fraction  will  he  multiplied  as 
many  times  as  there  are  units  in  the  divisor. 

Hence,  a  fraction  is  multiplied,  hy  dividing  its  denominator. 

Art.  135.  If  the  numerator  of  a  fraction  be  multiplied 
by  any  number,  its  value  (prop,  i,)  will  be  multiplied  by 
that  number ;  if  the  denominator  be  multiplied,  the  value 
(PROP.  Ill,)  will  be  divided  by  that  number. 

Hence,  if  both  terms  are  multiplied  by  the  same  number,  the 
increase  from  multiplying  the  numerator,  equals  the  decrease 
from  multiplying  the  denominator:  and  the  value  is  not  changed. 

Thus,  3X2^1    ^^,    3X3^^_ 
'  5X2      10  5X3      15 

Illustration. — Multiplying  both  terms  of  the  fraction  |  by  2, 
gives  -A,  in  which  the  parts  are  twice  as  many,  but  only  one-half 
the  size.     Multiplying  both  terms  of  ^  by  3,  gives  y^^;  three  times 
as  many  parts,  each  part  one-third  the  size.     Hence, 

Prop.  V.  Multiplying  hoth  terms  hy  the  same  numher, 
changes  its  form,  but  does  not  alter  its  value. 

Art.  136.  If  the  numerator  of  a  fraction  be  divided 
by  any  number,  its  value  (prop,  ii,)  will  be  divided  by 
that  number;  if  the  denominator  be  divided,  the  value 
(prop.  IV,)  will  be  multiplied  by  that  number. 

Hence,  if  both  terms  are  divided  by  the  same  number,  the 
decrease  from  dividing  the  numerator,  equals  the  increase  from 
dividing  the  denominator :  and  the  value  is  not  changed. 

Thus,  A- 2     A     ^^^    1^3^^ 
12  H- 2      6  12 -T- 3      4' 

Review. — 132.  "What  is  the  eflFect  of  dividing  the  numerator  of  a  frac- 
tion, without  changing  the  denominator?  133.  What  of  multiplying  the 
denominator  without  changing  the  numerator? 

134.  What  of  dividing  the  denominator,  without  changing  the  nume- 
rator?    135.  What  of  multiplying  both  terms  by  the  same  number? 


142  RAY'S   PRACTICAL    ARITHMETIC. 

Illustration. — Dividing  both  terms  of  (he  fraction  j^^^  by  2,  it 
gives  |;  in  which  there  are  07ie- half  a.s  many  parts,  but  each  part 
is  twice  the  size.  Dividing  both  terms  of  -^^  by  3,  gives  |,  one- 
third  as  many  parts,  each  part  three  times  the  size.     Hence, 

Prop.  VI.  Dividing  both  terms  by  the  same  number^  changes 
its  form^  but  does  not  alter  its  value. 

To  Teachers. — By  considering  the  numerator  a  dividend,  the  denomi- 
nator a  divisor,  and  the  value  of  the  fraction  the  quotient  (Art.  125),  the 
preceding  propositions  may  be  regarded  as  inferences  f n  m  Art.  57,  58,  59. 
This  short  method  is  not  best  adapted  to  young  pupils. 

Art.  137.  reduction  of  fractions 

Is  changing  their  form  without  altering  their  value. 

CASE   I. 
Art.  138.    To  reduce  a  fraction  to  its  lowest  terms. 

A  fraction  is  in  its  lowest  terms,  when  the  numerator 
and  denominator  are  prime  to  each  other.  Art.  110,  Def.  5. 

Thus,  i  is  in  its  lowest  terms,  while  ^J  is  not. 

1.  Reduce  -JJ  to  its  lowest  terms. 

Solution. — Since  the  value  of  a  first  operation. 

fraction   is   not  altered  by  dividing  24      ^)l'?       4 

both    terms    by    the    same    number,      2).3-7,=       iK  ^^  ^      "^' 
(Art.  136),  and,  as  two  is  a  common 
factor,  divide  both  terms  by  it;  the  fraction  then  becomes  j3 

Again,  since  3  is  a  factor  of  12  and  15,  divide  them  both  by  it; 

the  result,  -|.  can  not  be  reduced  lower. 

second  operation. 

Instead  of  dividing   by  2,  and   then  by  3,         a\^^ ^     a 

divide  at  once  by  6,  the  greatest  com.  divisor  ^ST)       5 

of  the  two  terms,  and  the  result  is  the  same. 

Solve  the  two  following  Examples  by  both  methods. 

Note. — All  subsequent  Exanijiles  having  a  star,  -X-,  arc  intended  to 
illustrate  the  principles  on  which  the  next  succeeding  rule  is  founded. 
The  pupil  should  solve  them  and  explain  the  operation,  referring,  at  the 
conclusion  of  the  exercise,  to  the  rule  which  follows. 


CO:\rMON    FRACTIONS.  143 

*2.  Reduce  ^f  to  its  lowest  terms.    .     .     .     Ans.  |. 
*3.  Reduce  |g  to  its  lowest  terms.    .     .     .     Ans.  |. 

Rule  for  Case  I. — Divide  the  numerator  and  denominator 
hy  any  common  factor ;  divide  the  resulting  fraction  in  the 
same  manner^  and  so  on  till  no  number  greater  than  1  will 
exactly  divide  both  terms. 

Or,  Divide  the  numerator  and  denominator  by  their  greatest 
common  divisor;  the  resulting  fraction  will  be  in  its  loicest  terms. 

Rem. — When  the  terms  of  a  fraction  are  small,  the  first  method  is  most 
convenient  ;  when  large,  the  second  method. 

REDUCE   TO  THEIR  LOWEST  TERMS, 

4. 
5. 

6. 

7. 
8. 
9. 

EXPRESS   IX   ITS   SIMPLEST   FORM, 

16.  The  quotient  of  391  divided  by  667.  Ans.  -|-|. 

17.  The  quotient  of  585  divided  by  1287.         Ans.  fj. 

18.  The  quotient  of  796  divided  by  1-4129.       Ans.  =4 


1-2 

T8- 

Ans. 

2 

10. 

126 

Ans. 

7 
1  I* 

30 
4.5- 

Ayis. 

2 

11. 

]  8  2 
T96- 

Ans. 

1  3 
14- 

JLO_ 

J  50* 

Ans. 

12. 

6  1  5 
^15- 

Ans. 

4  1 
61* 

4-2 

70- 

Ans. 

3 
5- 

13. 

87  3 
T067- 

Ans. 

9 
iT* 

96 
]  1  2" 

Ans. 

6 
7- 

14. 

777 
1998' 

Ans. 

7 
IS- 

60 
T25- 

Ans. 

1  2 
3?- 

15. 

909 
^323- 

Ans. 

9 
23- 

CASE    II 


Art.  139.    To  reduce  an  improper  fraction  to  a   whole 
or  mixed  number. 

1.  In    4   halves  (4)    of  an   apple,    how  many  apples? 
in  6  thirds  (f )  ?     in  8  fourths  (|)  ?     in  9  ?     in  \-  ? 

2.  In  8   pecks,  that   is,    in  |   of  a   bushel,   how  many 
bushels  ?     in  J  ?     in  Lf»  ?     in  y  ?     in  '-^  ? 

Review.— 136.  What  is  the  eflFect  of  dividing  both  terms  of  a  fraction 
by  the  same  number  ?     137.  What  is  Reduction  of  Fractions  ? 

138.  When  is  a  fraction  in  its  lowest  term>  ?  Give  an  example.  How 
is  a  fraction  reduced  to  its  lowest  terms,  Rule  ? 


144 


RAY'S  PRACTICAL  ARITHMETIC. 


3.  In  9  fourths  (|)  of  a  dollar,  how  many  dollars  ? 

Solution. — Since  4  fourths  make  one  dollar, 
there  are  as  many  dollars  as  there  are  times  4 
fourths  in  9  fourths;  that  is,  21  dollars. 

4.  Reduce  y  to  a  mixed  number. 


OPERATION. 

4)9 
Am.  ^'2\. 


Solution. — Since  5  fifths  make  1  (unit), 
there  will  be  as  many  ones  as  there  are  times  5 
in  17;  that  is,  3|. 

*5.  In  ^J  of  a  dollar,  how  many  dollars? 
^-6.  Reduce  r/  to  a  mixed  number. 


OPERATION. 

5)17 

Ans.  3t 

Ans.  2^\. 

Ans.     8|. 


Rule  for  Case  II. — Divide  the  numerator  by  the  denomi- 
nator :  the  quotient  will  be  the  whole  or  mixed  number. 

7.  In   -y   of  a  dollar,  how  many  dollars?  Ans.  13|. 

8.  In   Y    of  a  yard,    how  many  yards?  Ans.  IS^. 

9.  In  'g5  of  a  mile,     how  many  miles?  Ans.  15g. 
10.  In  y^'  of  a  day,      how  many  days?  Ans.  25-^i. 


n. 

12. 
13. 
14. 


REDUCE   TO    WHOLE    OR   MIXED   NUMBERS, 

Ans.    1.       15 


19 

775 
25  • 

17  1 
T-J  • 

5  0_9 
Tl  • 


Ans.  31. 
Ans.  14  A. 

1     -^r 

Ans.  46/-. 


16. 
17. 

18. 


6_4_3  7 

7  5  sn 
I'i'S  • 

3781 
T^    • 
1325 
TOT  • 


Ans.  2nil 
Ans.  eO-j-^^fij. 
Ans.  199. 
Ans.  13j^^-^. 


CASE    III. 

Art.  140.    To  reduce  a  whole  or  mixed  number  to  an 
improper  fraction. 

1.  In  2    apples,  how  many  halves  ?  In  3?  In  4? 

2.  In  2    apples,  how  many  thirds?  In  3?  In  4? 

3.  In  2    apples,  how  many  fourths?  In  3?  In  4? 

4.  In  2  J  apples,  how  many  halves  ?  In  3^  ?  In  4A  ? 

5.  In  2^  apples,  how  many  thirds  ?  In  2j  ?  In  33^  ? 


Review. — 188.  Why  is  the  value  of  a  fraction  not  altered  by  being 
reduced  to  its  lowest  terms?  189.  How  is  an  improper  fraction  reducod 
to  a  whole  or  mixod  number,  Rule? 


COMMON   FRACTIONS. 


145 


6.  In  5|  dollars,  how  many  fourths?     Or,  reduce  5|  to 
an  improper  fraction. 


SoLDTiON . — Since  there 
are  4  fourths  in  $1,  in  5  dollars 
there  are  5  times  as  many;  5 
times  4  fourths  are  20  fourths, 
and  20  fourths -f  3  fourths,= 
23  fourths.     Am.  2-3. 


5| 


OPERATIOX. 


20  =  fourths  in  5  dollars. 
3  =  fourths  in  fraction. 

23  =  fourths  in  5|. 


*7.  In  8|  apples,  how  many  fourths? 
*8.  Reduce  12|  to  an  improper  fraction. 


Ans.   3-5. 
Ans.   ^J. 


Rule  for  Case  III. — Multiply  the  ichole  number  by  tJie  de- 
nominator  of  the  fraction  ;  to  the  product  add  the  numerator., 
and  write  the  sum  ovei'  the  denominator. 

Rem. — The  analysis  of  question  6,  shows  that  the  whole  number 
is  really  the  multiplier,  and  the  denominator  the  multiplicand; 
but  the  result  will  be  the  same  (Art.  30),  if  the  denominator  be 
taken  as  the  multiplier. 


9.     In    5 j^Q  dollars,  how  many  tenths  ?      Ans. 

10.  In  15|    yards,     how  many  sixths  ?      Ans. 

11.  In  26.^1  days,      how  many  24:ths?       Ans. 


5  3 

TO- 
93 
6   • 
637 
24   • 


REDUCE   TO   IMPROPER   FRACTIONS, 


12. 

13. 
14. 
15. 


8i. 

311. 
465 


Ans. 
Ans. 
Ans. 
Ans. 


1 7 
2  ■ 

2_1 

4  ' 
i_8:2 
55  • 
373 


16. 
17. 

18. 
19. 


1  _9  9  9_ 
-^  1  0  0  0  • 


Ans. 
Ans. 
Ans. 
Ans. 


12360 
.t8  3     • 

1  999 
T(500' 

1  000 
7J     • 
1111 
TlT  • 


Art.  141.    To    reduce    a   whole    number   to   a  fraction 
having  a  given  denominator. 

1.  Reduce  3  to  a  fraction  whose  denominator  is  4. 


Solution. — Since  there  are  4 
fourths  in  1,  in  3  there  will  be  3 
times  4  fourths  =  12  fourths ;  and 
hence,  3  =  ^-^  Ans. 


OPERATION. 

4  =  fourths  in  1. 
3 


12  =  fourths  in  3. 


3dBk. 


10 


146  RAY'S   PRACTICAL    ARITHxMETIC. 

Hule. — Multiply  together  the  whole  number  and  the  denomi- 
nator;  beneath  the  product  write  the  denominator. 

2.  Reduce  4  to  a  Frae.  whose  denom'r  is  7.  Ans.    ^^. 

3.  Reduce  8  to  ninths Ans.    "y. 

4.  19  to  nineteenths Ans.  \y. 

5.  37  to  a  fraction  whose  denom.  is  23.       Ans.  %%^. 

CASE    IV. 
Art.  142.    To  reduce  compound  to  simple  fractions. 

1.  Reduce  J  of  |  to  a  simple  fraction. 
Analysis.— I  of  i  is  2  times  as  operation. 

much  as  \  of  |,  and  i  of  |  is  4  times         ^  of  -    ==  ^  ^  . 

as  much  as  1  of  i;  but  1  of  i  =  -Jg  3        5       3X5"~15 

(Art.   130);  and  hence,  1  of  f  =  4 

times  Jg  ^  ^%  (Art.  131),  and  f  of  J  =  2  times  j%  =  j%. 

In  this  operation,  the  numerators  are  multiplied  together,  as 
also  are  the  denominators. 

*2.  Reduce  |  of  |  to  a  simple  fraction.         Ans.  ^2. 
*3.  Reduce  |  of  J  to  a  simple  fraction.         Ans.  |^. 

Rule  for  Case  IV. — MuUiphj  the  numerators  together  for  a 
new  numerator^  and  the  denominators  together  for  a  new 
denominator. 

If  mixed  numbers  occur,  reduce  them  to  improper  fractions. 

4.  Reduce  A  of  j  of  2|  to  a  simple  fraction. 

S0L.-2|  =  V,  and  A  of  f  of  U  ■=  |§- 

5.  Reduce  j'y  of  |  to  a  simple  fraction.         Ans.    ^^. 

6.  I  of  I  to  a  simple  fraction.  Ans.    ^§. 

7.  f  of  I  of  1^  to  a  simple  fraction.  Ans.  j^. 

Review. — 140.  IIow  is  a  mixed  number  reduced  to  an  improper  fr« 
tion,  Rule  ?    141.  IIow  is  a  whole  number  reduced  to  a  fraction  having  a 
given  denominator,  Rule  ? 


COMMON    FRACTIONS.  147 

8.  Reduce  §  of  |  of  |  to  a  Bimple  fraction. 

Solution. — After    indicating    the  operation. 

operation,  the  numerator  of  the  result  1         1 

will  be  2X3X4;  the  denominator,  2       ?>        ^       2 

3X4X5.  '|^^^'5"~5^"^'- 

The  value  of  a  fraction  not  being         j         j 
altered  by  dividing  both  terms  by  the 

same  number  (Art.   136),  Cancel  the  factors  (8  and  4,)  common  to 
both  terms. 

As  3  =  3X1,  and  4  =  4X1,  the  factors  1  and  1  will  remain 
after  canceling  3  and  4.  Hence,  the  products  of  the  remaining 
factors  are  2x1X1,  and  1X1X5,  which  give  the  terms  of  the 
required  fraction  in  its  simplest  form. 

*  9.  Reduce  J  of  |  of  |  to  a  simple  fraction.  Avs.  ^-^. 
||e     *10.  Reduce  |  of  ^  of  J  to  a  simple  fraction.  Am.    |. 

Art.  143.  Hence,  to  reduce  compound  to  simple 
fractions  by  Cancellation, 

Indicate  the  operation ;  cancel  all  the  factors  common  to  both 
terms^  and  multiply  together  the  factors  remaining  in  each. 

Rem. — As  all  the  factors  common  to  both  terras  are  canceled  by 
the  operation,  the  result  will  be  in  its  simplest  form. 

11.  Reduce  |  of  4  of  /tj  of  ]|  to  a  simple  fraction. 

Solution. — First,   cancel    the  operatiOiV. 

:   factors  3  and  4  in  the  numera-  2 

[   tor,  and  12  in  the  denominator,        Q        A        J        1^       2 
I  as  4X3  =  12.  ^X  ^X   '     Xq^=25^^'' 

Since  9  is  a  factor  of  18,  can-  }^       A^       PP      ^*^ 

eel  the  factor  9  i  n  bof  h  terms,  and 

write  the  remaining  factor,  2,  above  18 ;  as  7  is  a  factor  of  85,  can- 
cel the  factor  7  in  both  terms,  and  write  the  remaining  fac- 
tor, 5,  below  35.    Then  multiply  the  remaining  factors  as  before. 

Review. — 142.  How  are  compound  reduced  to  simple  fractions,  Rule  ? 
143.  How  reduced  by  Cancellation  ?  Why  is  the  value  of  the  fraction  not 
altered  ?     Rem.  Why  is  the  result  in  its  lowest  terms  ? 


148 


RAYS  PRACTICAL  ARITHMETIC. 


REDUCE 

TO     SIMPLE 

FRACTIONS, 

12. 

I  of|of    i. 

Ans.  i. 

16. 

.s.of.^ofl?.    Ans.  I 

13. 

i  off  of  11. 

Ans.  -i. 

17. 

I  of  |of  Jof  5.  ^«s.  1. 

14. 

loffofli^ 

Ans.  1. 

18. 

>  off  off  of  1  of 

15. 

3   off   of  1|. 

Ans.  2. 

loflof,-^^.     ^«s.,V 

For  method  of  reducing  complex  to  simple  fractions,  see  page  167. 

CASE   V. 
Art.  144.    To  reduce  fractions  of  different  denominators 
(a  equivalent  fractions  having  a  common  denominator. 

1.  Reduce  A,  f ,  and  |  to  a  common  denominator. 


3>    3 

Solution. — The  value  of  a 
fraction  not  being  altered  by 
multiplying  both  terms  by  the 
same  number  (Art.  135),  multiply 
the  numerator  and  denominator 
of  each  by  the  denominators  of 
the  other  fractions ;  this  will  ren- 
der the  new  denominator  of  each 
the  same;  since,  in  each  case,  it 
will  consist  of  the  product  of  the 
same  numbers,  that  is,  of  all  the  denominators. 


OPERATION. 

1X3X4_12  new  numer. 
2X3X4      24  new  denom. 


2X2X4. 
3X2X4" 


.1^6  new  numer. 
24  new  denom. 


3X2X3_18  new  numer. 
4X2X3      24  new  denom. 


*2.  Reduce  ?  and  |  to  a  com.  denom.     Ans.  ^§,  j§. 
*3.     ^,  |,  and  |  to  a  com.  denom.       Ans 


3  0      3  fi      7  5 

^0'  yO»  '90' 


Bule  for  Case  V. — Multiply   both   terms  of  each  fraction 
by  the  product  of  all  the  denominators  except  its  own. 

Note. — First  reduce  compound  to  simple  fractions,  and  whole 
or  mixed  numbers  to  improper  fractions. 

4.  Reduce  A?  3,  and  Z  to  a  common  denominator. 


Suggestion. — Since  the  denomi- 
nator of  each  new  fraction  consists 
of  the  product  of  the  same  num- 
bers, (all  the  denominators  of  the 
given  fractions,)  we  multiply  them 
together  but  once. 


OPRRATTON. 

1X3X5  =  15  1st  num. 
4X2X5=40  2d  num. 
7X2X3=42  3d  num. 

2X3X5=30  denom. 


COiMMON   FRACTIOr^S. 


149 


Observe,  that,  in  each  case,  the  result  is  obtained  by  multiplying 
the  numerator  and  denominator  by  the  same  number. 


REDUCE   TO   COM.    DENOMINATORS, 


5. 

2,  -4,  ana  g. 

Ans.  |o, 

30 

405 

il 

6. 

hh  andf 

^^s.  M, 

8 
245 

A- 

7. 

1,  1,  and  f . 

Ans.  3  5, 

42 

7  05 

le- 

8. 

12     3     anrl    4 
2J   35  4'  *^    ^   5' 

J.71S.     rj^^Q,        JTj^0, 

90 

1205 

T20* 

9. 

i  1,  1,  and  f . 

Ant     56^        335 
-^?iS.  gj^,     840» 

3£fi 
8405 

525 
840' 

10. 

A>  ^'  and  ^3. 

J/Mo       6  4  6 

572 
100  15 

_6_1_6_ 
lOOl* 

11. 

i  off,  2J,  and  3. 

Ans.    3^4, 

5^ 
245 

72 
24* 

12. 

1  of  1,  and  i  of  1 

off   of  If.             ^715. 

35 
"5  6  5 

8 
56- 

Art.  145.  When  the  given  fractions  are  expressed  in 
small  numbers,  and  the  denominator  of  either  fraction  is 
a  multiple  of  the  denominators  of  the  others,  reduce  them 
to  a  common  denominator  ;    thus, 

Multiply  both  terms  of  each  fraction  by  such  a  number  as 
will  render  its  denominator  the  same  as  the  largest  denomi- 
nator; obtain  this  number  by  dividing  the  largest  denominator 
by  the  denominator  of  the  fraction  to  be  reduced. 

OPERATION. 


1.  Reduce  i  and  i  to  a  com.  denom. 

Solution. — Since  the  largest  denom.,  6,  is  a 
multiple  of  2,  multiplying  both  terms  of  ^  by  |=3, 
reduces  it  to  |. 


Am.  I  and  |. 


1X3 
2X3' 

1    ^ 
6 


3 

'6 

1 
6 


By  similar  process,  Reduce  to  Com.  Denominators, 


EXAMPLES. 

2.  i  and  I    = 

3  111    = 

*^*  25     45      8 

^'  35     25      g 

5  1     2    _3_   _ 

*'•  25     65    10 


ANSWERS. 

2 

45 

4        2 
55      H5 
o 
55 


2 
4* 

1 

5- 
5 


3 

g5 

5         4         3 
TO'  T05  TU 


EXAMPLES. 


ANSWERS. 


6. 

7. 
8. 
9. 


1 
35 

1 

25 

45 

2 
35 


5 
55 

4 

75 

5 

85 

3 

45 


7_ 
12 

_9_ 
14 

1  1 

Tg 
11 
12 


4 
1^1 

7_ 
145 

12 

Tg5 

8 
T2' 


10 
T55 

8_ 

145 

10 
165 

9_ 

125 


7 
T2- 

_9_ 

14* 

1  1 

Tg- 

11 

12- 


Eeview. — 144.  How  are  two  or  more  fractions  reduced  to  a  common 
denominator?  Why  is  the  value  of  each  fraction  not  altered?  Why 
does  this  operation  render  the  new  denominator  of  each  the  same  ? 


150 


RAYS    PRACTICAL    ARITHMETIC. 


CASE    VI. 


Art.  146.    To  reduce  fractions  of  different  denominators^ 
to  equivalent  fractions  having  the  least  com.  denominator. 

1.  Reduce  ^,  |,  and  |  to  the  least  com.  denominator. 


OPERATION. 

2)2     3     4 
1      3     2 

2X3X2  =  12,  least  com.  mul. 

4112 
3 


2)12 

3)12 

6 

4 

1X6 

6   >| 

2X6 

12 

2X4 
3X4 

8 
~12 

3X3 

9 

4X3 

~12  J 

Solution. — Since  multiplying 
both  terms  of  m  fraction  by  the 
same  number,  does  not  alter  its 
value,  a  fraction  may  be  reduced 
to  another  whose  denominator  is 
any  multiple  of  the  denominator 
of  the  given  fraction. 

Thus,  ^  may  be  reduced  to  a 
fraction,  whose  denominator  is 
either  4,  G,  8,  10,  12,  14,  16,  &c. 

And,  I  may  be  reduced  to  a 
fraction,    whose    denominator  is  -^''^^  =  —    L  Arts. 

either  6,  9,  12,  15,  18,  21,  &c. 

And,  I  may  be  reduced  to  a 
fraction  whose  denominator  is 
either  8,  12,  16,  20,  24,  &c. 

We  find  12  to  be  the  least  denominator  commmi  to  all  of  them. 
These  denominators  being  multiples  of  the  denominators  of  the  given 
fractions,  it  follows,  that  12,  their  least  common  multiple,  is  tiie  least 
com.  denominator  to  which  the  fractions  can  be  reduced. 

It  now  remains  to  reduce  the  fractions  to  TWELFTIIS. 

Thus,  -^  will  be  reduced  to  twelfths  by  multiplying  both  of  its 
terms  by  6,  which  is  the  quotient  of  the  L.  C.  M  ,  12,  divided  by  2. 

And,  I  will  be  reduced  to  twelfths,  by  multiplying  both  of  it? 
terms  by  4,  the  quotient  of  12  divided  by  3. 

And,  I  will  be  reduced  to  twelfths,  by  multiplying  both  of  its 
terms  by  3,  the  quotient  of  12  divided  by  4. 

Review. — 145.  When   the   denominator  of  one  of  the  fractions  ia  a 
multiple  of  the  others,  how  reduce  them  to  a  com.  denominator  ?     IIow  is  j 
the  multiplier  of  each  fnujtion  obtained  ? 

14^'.  What  are  the  denominators  of  the  fractions  to  which  ono-half  may] 
be  reduced?     Two-thirdd  ?     Three-fourths? 


COMMON    FRACTIONS.  151 

With  regard  to  this  Operation,  the  pupil  will  notice, 

1st.  When  a  fraction  is  in  its  lowest  terms,  the  denominator 
of  any  fraction  to  which  it  can  be  reduced  must  be  a  multiple 
of  the  denominator  of  the  given  fraction  :  hence, 

Any  denominator  common  to  two  or  more  fractions  must  be 
a  common  multiple  of  their  denominators :  therefore, 

The  least  common  multiple  of  their  denominators,  is  the  least 
com.  denominator  to  which  two  or  more  fractions  can  be  reduced. 

2d.  The  values  of  the  fractions  are  not  altered,  for  both  terms 
are  multiplied  by  the  same  number  (Art.  135.). 

REDUCE   TO   THEIR   LEAST   COMMON   DENOMINATORS, 

*2.     \  and  I Ans.  -f^  and  -{J. 

^ o.     T7j  "-J J  anci  -jj.    «     .     »     •     .  .A71S.  Ygj  Tft)  aDQ  ■j'^* 

Rule  for  Case  VI. — Find  the  least  common  multiple  of  the 
denominators  of  the  given  fractions,  and  multiply  both  terms 
of  each  fraction  by  the  quotient  of  the  least  common  multiple^ 
divided  by  the  denominator  of  the  fraction. 

Notes. — 1.  Before  commencing  the  operation,  each  fraction  must 
be  in  its  lowest  terms. 

2.  Compound  must  be  reduced  to  simple  fractions,  and  whole  or 
mixed  numbers  to  improper  fractions. 

3.  After  the  pupil  is  well  acquainted  with  the  nature  of  the 
operation,  the  multiplication  of  the  denominators  may  be  omitted, 
as  the  new  denominators  will  be  equal  to  the  L.  C.  M. 

4.  The  object  of  reducing  fractions  to  a  common  denominator,  is 
to  prepare  them  for  addition  and  subtraction,  which  can  be  per- 
formed only  when  the  numbers  are  of  the  same  unit  value. 

Eeview. — 146.  What  is  the  least  denominator  common  to  the  fractions 
whose  denominators  are  2,  3,  and  4  ?  How  reduce  one-half  to  twelfths  ? 
Two-thirds  ?  Three-fourths  ?  Why  is  the  L.  C.  M.  of  the  denominators, 
the  least  common  denominator?  Why  are  the  values  of  fractions  not 
altered  by  the  operation  ?     What  is  the  Rule  for  Case  VI  ? 


152 


RAY'S    PRACTICAL    ARITHMETIC. 


REDUCE   TO   THEIR   LEAST   COM.    DENOMINATOR, 


EXAMPLES. 


4. 
5. 
6. 

10. 
11. 
12. 


1 

1 


1 

3' 

1 


3'    g' 
1      3 

2'   4' 


ANSWERS. 

6 
T5' 

6 
Tg' 


20' 


A'  T2- 
__3      _2 

ig'  Ig- 
15  16 
2U'  5(}' 


I   8.    i 
i   9. 


EXAMPLES. 

3  4     _9 

8'  5'    lO 

2  3     7 
3'  4'   S 

3  5      5 
4'  g'    ^ 


ANSWERS. 


15 

40' 

16 

24' 

54 

72' 


32 

4  0' 

24' 
45 

72' 


36 
4  0- 
21 
24- 

40. 
72- 


2 
7' 

i. 

3 
4' 


JL        9_       1  1 
14'     2i'     2g- 

2       3       4       5 
3'      4'      5'      6" 

A  of  ^     J-Q 

2  "^     4'      12' 


(See  Note  1.)       Ans. 


Am.  ^fi 


?2 


J.IIS. 


,8^      10       12       11 

2g'     28'     28'     2g- 

40       45       48       5Q 

6  0'      eU'.    6  0'      60- 

_9,0_       1I}5       100      288 

120'      120'      120'     T20' 


60' 


1&^^  For  additional  problems,  see  Ray's  Test  Examples. 

Art.  147.  addition  of  fractions 

Is  the  process  of  uniting  two  or  more  fractional  numbers. 

1.  What  is  the  sum  of  I  and  f  and  |? 

Solution. — Since  the  denominators  are  the  same,  the  numerators 
express  parts  of  the  same  size :  therefore,  add 

1  Jifth,    )  (  1  cent, 

2  Jifihs,  >  as  you  would  add  <  2  cents, 

3  fifths,  )  -  (  3  cents, 

The  sum  is  6  fifths  (|)  in  one  case  ;  0  ce7i^5  in  the  other. 

Hence,  to  add  fractions  having  a  Com.  Denominator, 
find  the  sum  of  the  numerators ;  write  the  result  over  the 
denominator.  • 

EXAMPLES    FOR   MENTAL    SOLUTION. 


2. 

Add 

1 

4' 

I 

and 

3. 

1 

2 

3 

5) 

6» 

5' 

4. 

1 

2 

5 

7) 

7? 

^' 

3 

4* 

4 

o' 


5.  Add 
6. 

7  3        7      _8      JL     IQ 

*•  T2»  T2>    l2»    12»    12- 


2        3.^ 
g»      8»      5' 
1  2  3 

TO'     lO'    lO» 


Kkview, — 14r>.  Note  1.  Before  commencing  the  operation,  what  i* 
required  ?     2.  If  there  are  compound  fractions  or  mixed  numbers? 

Note  3.  What  may  bo  omitted  ?  Why?  4.  What  is  the  object  in  reducing 
fractions  to  a  common  denominator?  147.  What  is  Addition  of  Friwjtions? 
IIow  aUd  fractions  having  a  common  denominator  ?     Why  ? 


I 


COMMON   FRACTIONS.  153 

Art.  148.  1.  What  is  the  sum  of  h  and  I  ? 

Solution. — Since  the  denominators  are  dif-  operation. 

ferent,  the  numerators  do  not  express  things  h^^  B 

of  the  same  unit  value,   and    they  can  not  be  1  =  i 

added  together.    The  sum  of  1  half  and  1  third,  312 5    a 

is  neither  2  halves  nor  2  thirds.     But,  reduced 

to  a  common  denominator,   (Art.  144),   1  half  =3  sixths,  and  1 

third  =  2  sixths  ;    their  sum  is  5  sixths  (|). 

*2.  Add  I  and  I Ans.  f^. 


* 


3.  Add  I  and  | Ans.  JJ  =  ljo- 


Rule  for  Addition. — Reduce  the  fractions  to  a  common 
denominator  ;  add  their  numerators  together,  and  place  the  sum 
over  the  common  denominator. 

Notes. — 1.  Reduce  compound  to  simple  fractions,  and  each  frac- 
tion to  its  lowest  terms,  before  commencing  the  operation. 

2.  Mixed  numbers  may  be  reduced  to  improper  fractions,  and 
then  added ;  or  the  fractions  and  whole  numbers  may  be  added 
separately,  then  united. 

3.  After  adding,  reduce  the  result  to  its  lowest  terms.     Art.  138. 

4.  What  is  the  sum  of  |,  |,  and  |? 

OPERATION. 

2)3   4   6-    12 


I 


P 


3)3    2    3     12 
12    1     12 


3  =  4,  and  4X2  = 

4  =  3,  and  3X3=    9  [  ^^^^^ 
6  =  2,  and2x5=10 


2X3X2  =  12,  Least  Com.  Mult.     -^^'*'-   1  ^i  "  ^  1 2        ^4' 

Suggestion. — In  reducing  the  fractions  to  their  least  common 
denominator,  omit  writing  the  denominator  beneath,  until  the  sum 
of  the  numerators  is  obtained. 

5.  Add     g4^,      i?,      §a       ff,    and    Jf.  Ans.  2. 

fi       Arlrl        IL        _L9_        _29_        10  1  J  7,0     11 

'-'•    xi-UU      i4di       idd)       AAA')       144 xi.n!>.   ±Q. 

'_83_ 
125- 


b\. 

57' 

57' 

1 1_ 

144' 

19 
144' 

_29 
144' 

9  9  9_ 

rooo' 

888 
lOOU' 

777 
1000 

7       XAA      999_     _8_88       JJJ__  A^j^     9_83_ 


Review. — 148.  Why  can  not  1-half  and  1-third  be  added,  without  re- 
ducing them  to  the  same  denominator  ?     What  the  rule  for  addition  ? 


154 


RAY'S   PRACTICAL    ARITHMETIC. 


8.  Add 

9.  Add 
10.  Add 

14.  Add 

15.  Add 
IG.  Add 

17.  Add 

18.  Add 

19.  Add 
20  Add 

J6@°"  For 


5      1 

5'    1  u 


I  8' 

Ans.  l-Jf. 


12'     135     ]4»  '*'""    1  5* 


I  and  J..  Ans.  1/^.  11.  Add  J,  |,  ]4.    ^^is.  2^\ 
^«s.    12.  12.  Add  i,  1,  y2_.    ^ns.  I^A 
13.  Addi,li,l,i.AnsA^ 

Ayis.  jg.jo- 

2^  and  3» 

I,"  7A,  and  8| 

16i,  12|,  8f,  and  2|.  .  . 
I  of  I,  and  A  of  I  of  j%r.  . 
i  of  1  1     1   of  2 »     1  1  of  1     '7 


55     1  gJ    505     I4l)5    ^"^    ^ZHWO' 


Ans.  5 "3. 

1  o 

Ans.    |. 
^ns.2J33. 

^4  MS.       1. 


J 


additional  problems,  see  Rays  Test  Examples. 


Art.   149.   SUBTRACTION    OF    FRACTIONS 

Is  the  process  of  finding  the  difi"erence  between  two 
fractional  numbers. 

1.  What  is  the  difference  between  ^  and  ^  ? 

Solution. — Since  the  denominators  are  the  same,  the  numerators 
express  parts  of  the  same  size:  therefore,  subtract  2  sevenths 
from  5  sevenths  as  you  would  2  cents  from  5  cents. 

Thus,  5  sevenths^  5  cents, 

2  sevenths,  2  cents, 

Difference  3  sevenths  (|)  in  one  case ;  3  cents  in  the  other. 

Hence,  to  find  the  difference  between  two  fractions 
having  a  common  denominator, 

Find  the  difference  between  their  numerators,  and  write  the 
result  over  the  common  denominator. 

QUESTIONS   FOR   MENTAL   SOLUTION. 

2.  What  is  the  difference  between  \  and  |  ?     |  and  J  ? 
f  and§?  /,-and-,35?     yandj?     Uandf? 


Review. — 1-48.  Note.  If  there  arc  Compound  Fractions,  what  is  re- 
quired ?  What  if  each  fraction  is  not  in  ita  lowoat  terms  ?  How  are 
mixed  numbers  added  ? 


COMMON   FRACTIONS.  155 

Art.   150.  1.  Find  the  difference  between  i  and  r.. 

Solution. — Since    the    denominators    are  operation. 

different,    the    numerators    do   not    express  f^^e 

things  of  the  same  unit  value:  hence,  one  can  A  =^4 

not  be  subtracted  from  the  other.   Art.  25. 

Thus,  the  difference  between  1  half  and  2         g      6  —  6  ^ns. 
thirds  is  neither  1  half  nor  1  third;  but,  reduce 

them  to  a  common  denominator  (Art.  144),  and  2  thirds  =  4  sixths, 
and  1  half =3  sixths;    their  difference  being  1  sixth  (^). 

*2.  The  difference  between   *  and  |.  Ans.  -jL. 

*3.  Between  ^  and  4 Ans.  j^^. 

Rule  for  Subtraction. — Reduce  the  fractions  to  a  common 
denominator,  Jind  the  difference  of  their  numerators,  and  place 
it  over  the  common  denominator. 

Note. — Reduce  compound  to  simple  fractions,  and  each  fraction 
to  its  lowest  terms,  before  commencing  the  operation.  After  sub- 
tracting, reduce  the  result  to  its  lowest  terms. 

4.  What  is  the  difference  between  |  and  ^q  ? 

OPERATION. 

6  =  ^X3         30-f-    6  =  5,  and5x5  =  25|     New 
10  =  2X5         30---10  =  3,  and  3X  3=    9  J  Nume'rs. 

3X2X5  =  30,  Least  Cora.  Mult.  ^«s.]§  =  f^. 

5.  8  _  5   ^  ^^.,,     1.       10.      I  —  A  =  Ans.  ^\ 


"•  tI       T6  —  A.ns.     2- 

7.  I         -J   =  An^.  f^. 

ft  3    1    —  J„o    _7_ 

9.  4          ?  =  Ans.  3g. 


11.      5   _   3   ^  ^i„.s 


1 

7 

I 
1  I 

5 
6 

3      . 

5 
9 

1    — 

4 
1  5 

1 

8 
Z  1 

3     — 

14 

77* 
11 


12.      f  -  J  =  Ans.  j\. 
1-i.     A  —  A  =  ^«s. 


1 

6* 

1 

6* 


L  Rem. — In  finding  the  difference  between  mixed  numbers,  either 
reduce  them.to  improper  fractions,  and  to  a  common  denominator, 
and  then  make  the  subtraction;  or,  find  the  difference  between 
the  whole  numbers  and  the  fractions  separately. 

B  Review. — 149.  What  is  subtraction  of  fractions  ?  How  find  the  dif- 
ference between  two  fractions  having  a  com.  denominator  ?     Why  ? 


156 


RAY'S    PRACTICAL   ARITHMETIC. 


15.  From  3^  subtract  1|. 

Suggestion. — As  4  sixths  can 
not  be  taken  from  3  sixths, 
borrow  1  from  3,  reduce  it  to 
sixths,  add  them  to  the  3  sixths, 
making 9  sixths;  then  subtract. 

16.  From  5  subtract  |. 


7  —  21 
2—   g 


OPERATION. 

Or,  3.1 


5  — 
3 


10 

1 1 


n 


=  15 


15. 


Suggestion. — The    subtraction 


OPERATION. 

""^y    5=15        Or  5 
be  made  by  reducing  5  to  thirds;  or,  by  ^  '     o 

borrowing  1  from  5,  and  reducing  it  to  ^  •* 


thirds,  as  in  the  above  example. 


Ans.K^=4:].   4.1 


17. 

43. 

-2i=     21. 

23. 

11- 

_45              _      g4 

18. 

8J- 

-35=     4|. 

24. 

8- 

-3-3.       =4ifl 

19. 

5r 

41  —     1 1 

25. 

foflO- 

-§of6    =      2. 

20. 

n- 

43  __  2ii 

26. 

^of|- 

-  '  of  5             JL 

3  '^^   U      ?::• 

21. 

5   ■ 

11    —        31 

27. 

14J- 

-5  0fl9=l/j. 

22. 

8   - 

3                 74 

28. 

|of72- 

-3of21  =  14^. 

0                                                                   0 

For  additional  problems,  see  Ray's  Test  Examples. 


Art.  151.  MULTIPLICATION    OF    FRACTIONS. 

Case  I.    To  multiply  a  fraction  by  a  whole  number. 

This  operation  consists  in  taking  the  fraction  as  many 
times  as  there  are  units  in  the  multiplier. 

1.  If  1  apple  cost  1  of  a  cent,  what  cost  3  apples? 

Solution. —  Three  apples  cost  3  times  as  much  as  one;  that 
is,  J  taken  3  times :  i+^+J==  .|X3^f.     (Art.  131.) 

2.  If  1  lemon  cost  I  of  a  cent,  what  cost  4  lemons  ? 
Ans.  I  taken  4  times ;  |  +  3_|-|+|  =  | x 4=  i5*=2|. 

Review. — l.'O.  Why  can  not  one-half  be  taken  from  two-thirds  without 
reducing  to  the  same  denominator  ?     "What  the  rule  for  subtraction  ? 

150.  Note.  What  is  required  if  there  are  compound  fractions?  What 
if  each  fniction  is  not  in  its  lowest  terms?  Rem,  How  find  the  difference 
between  two  mixed  numbers  ?     Between  a  whole  number  and  a  fraction  ? 


COMMON   FRACTIONS.  157 

Another  Solution. — Since  3  fifths  express  three  parts  of  the 
same  size,  we  may  multiply  Z  fifths,  as  we  would  multiply  3  cents. 

Thus,  SJifths  And,  3  cents 

multiplied  by       4  multiplied  by      4 

the  product  is   12  ffths  (L2).         the  product  is  ]2  cents. 

Hence,    To   multiply   a  fraction    hy   a   whole   number^ 
multiply  its  numerator, 

EXAMPLES   FOR   MENTAL   SOLUTION. 


3. 

J  X    3  -     Ans.  |. 

7. 

1  X    6  =  Ans.  24. 

4. 

J  X    4  =  Ans.  IJ. 

8. 

f  X    8  —  Ans.  5J. 

5. 

f  X    6  -    Ans.  4. 

9. 

2  X     6  —  Ans.  4|. 

6. 

1  X  10  =    Ans.  6. 

10. 

1  X  10  =  Ans.  6]. 

11.  What  is  the  product  of  -J  multiplied  by  4? 

Solution. — Since    a    fraction 

is  multiplied  by  multiplying  its  '  operation. 

numerator,    (Art.    181,)     multi-        b'^'*^^  S  ^^5 ^^^2   Ans. 
ply    7    (eighths)  by   4,  and  the  or, 

product  is  28  (eighths);  which,        T'X4=:     ^  =  ^^3-t. 
reduced,  equals  31. 

Or,  since  a  fraction  is  multiplied  by  dividing  its  denominator, 
(Art.  134,)  if  the  denominator,  8,  is  divided  by  4,  the  resuL 
is  1=^3^,  the  same  as  by  the  first  method. 

*12.    2X3   =  Ans.  21.  I  *13.    -r'':,  X  4  =  Ans.  2|. 

Rule  for  Case  I. — 1st.  Multiply  the  numerator  of  the 
fraction  hy  the  ivhole  number,  and  under  the  product  write 
the  denominator. 

Or,  2d.  Divide  the  denominator  by  the  whole  number,  when 
it  can  be  done  without  a  remainder ;  over  the  quotient  ivriie 
the  numerator. 

14.    14X12  =  ^Tjs.  7f|.    I   15.    22_x7  =  Ans.  I. 

Note. — When  the  denominator  of  the  fraction,  and  the  whole 
number  contain  common  factors,  employ  Cancellation. 


158 


RAY'S   PRACTICAL    ARITHMETIC. 


16.     /^X 


M' 


3y. 


19. 
20. 
21. 


9  =  Ans. 

17.  §1  X    8=  Ans. 

18.  j%  X21=  A71S.  13.}. 

SuG. — In  multiplying  a  mixed  by  a 
■whole  number,  multiply  the  fractional 
part  and  the  whole  number  separately, 
then  unite  their  products;  or,  reduce 
the  mixed  number  to  an  improper  frac- 
tion, then  multiply. 


liXlO=    Ans.  7\. 

f  X    9=    Ans.     5. 

3]  X    4=  (Seesug.) 


opehation. 


4 


22.    18|  X  8  =  Ans. 


23. 


16§  X  3 


Ans. 


150. 
50. 


24. 
25. 


Ans.  13  A. 
Or,  31  =  LO  ; 

107  X  7  =  Ans.  75|. 
10|X9=  Ans.  97A. 


CASE   II. 
Art.  152.    To  multiply  a  whole  number  hy  a  fraction. 

Multiplying  by  a  whole  number,  is  taking  the  multiplicand 
as  many  times  as  there  are  units  in  the  multiplier. 

Multiplying  by  a  fraction,  or  part  of  a  unit,  is  taking  a  part 
of  the  multiplicand.     Therefore, 

Multiplying  by  -^,  is  taking  1  half  the  multiplicand. 

Multiplying  by  |,  is  taking  1  third  of  the  multiplicand. 

Multiplying  by  |,  is  taking  2  thirds  of  the  multiplicand,  &c. 

Hence,  To  multiply  by  a  fraction,  is  to  take  such  a  pari  of 
the  multiplicand,  as  the  mulliplier  is  part  of  a  unit. 

1.  At  12  cts.  a  yard,  what  cost  J  of  a  yard  of  ribbon? 

Solution. — If  1  yard  cost  12  cents,  1  third  of  a  yard  will  cost 
1  third  as  much,  that  is,  (Art.  54,)  12  Xg=4  cents.     Ans. 

2.  At  12  cts.  a  yard,  what  cost  ~  of  a  yard  of  ribbon  ? 

Solution. —  7'wo-thirds  will  cost  ttvice  as  much  as  orj^-third; 
but,  1  third  costs  4  cents,  hence,  2  thirds  cost  4X2  =  8  cents, 
that  is,  (Art.  54,)  12Xf  =  8  cents.     Ans. 

Rkview. — 151.  In  what  does  the  multiplication  of  a  fraction  by  a 
whole  number  consist  ?  What  is  the  first  method  ?  The  second  method  ? 

151.  Note.  When  the  denominator  and  whole  number  have  common  fac- 
tors, how  shorten  the  process  ?   ILtw  multiply  a  mixed  by  a  whole  number  ? 


i 


COMMON   FRACTIONS. 


159 


3.  "What  is  the  product  of  20  multiplied  by  |  ? 

Solution.  —  Three-lburths    are    3    times    1    fourth;    1    fourth 
of  20  is  5,  and  3  times  5  are  15:  Ans. 

Or,  1  of  20  is  2-0  ;  and  3  times  ^  are  2J)  x  3  =  6J)  =  15, 
EXAMPLES   FOR   MENTAL    SOLUTION. 


4. 

8  X  1          Ans. 

6. 

8. 

16X1- 

Ans.  14. 

5. 

12  X  f  —     Ans. 

8. 

9. 

5X  |- 

Ans.  3|. 

6. 

10  X  1  —     Ans. 

4. 

10. 

7Xf  = 

Ans.  4|. 

7. 

14  X  5  —     Ans. 

10. 

11. 

8X|  = 

Ans.  44 

Rule  for  Case  II. — 1st.  Divide  the  whole  number  by  the 
denominator  of  the  fraction  :  multiply  the  quotient  by  the 
numerator. 

Or,  2d.  Multiply  the  whole  number  by  the  numerator  of  the 
fraction^  and  divide  the  product  by  the  denominator. 

Rem. — 1.  The  2d  rule  is  best,  when  the  denominator  of  the 
fraction  is  not  a  factor  of  the  whole  number. 

2,  Since  the  product  of  two  numbers  is  the  same,  whichever  is 
the  multiplier,  (Art.  30,)  the  examples  in  this  and  the  preceding 
case  may  be  performed  by  the  same  rule. 

3.  When  the  multiplier  is  a  whole  number  greater  than  1,  the 
multiplicand  is  increased ;  when  it  is  a  proper  fraction,  the  mul- 
tiplicand is  decreased^  the  product  being  the  same  part  of  the 
multiplicand,  that  the  multiplier  is  of  unity. 

"      4.  Multiplying  by  a  fraction  always  involves  division.     Thus, 
multiplying  by  one-third,  is  the  same  as  dividing  by  three. 


12. 

28  X 

4  —  Ans.  16. 

15. 

31  X  1  —  Ans.  202. 

13. 

36  X 

I  -  Ans.  28. 

16. 

29  X  1  =  Ans.  21|. 

14. 

50  X 

^Q  —  Ajis.  45. 

17. 

37  X  4  =  Ans.  29-?. 

Review. — 152.  In  what  does  multiplication  by  a  whole  number  consist  ? 
In  what  the  multiplication  by  a  fraction,  or  part  of  a  unit  ?  What  is 
multiplying  by  one-half?    By  one-third  ?     Two-thirds? 

152.  In  multiplying  by  a  fraction,  what  part  of  the  multiplicand  is  taken  ? 
What  is  Rule  for  CiJse  II,  1st  ?    2d  Rule  ?  .  Rem.  1.  Which  is  best  ? 


160 


RAY'S   PRACTICAL   ARITHMETIC. 


18.  Multiply  8  by  3§. 

In  multiplying  a  whole  by  a 
mixed  number,  multiply  by  the 
integer  and  the  fraction  sepa- 
rately, then  unite  their  pro- 
ducts; or,  reduce  the  mixed 
number  to  an  improper  frac- 
tion, then  multiply. 

19.  25X8|=^7JS.    215. 

20.  45X61=  ^ns.  281  i. 


OPERATION. 

8 

Or,  3|-V 

^ 

8XV     %^     291. 

24 

^-- 

=8X| 

29] 

Ans. 

21. 

55X9|— ^»s.  518| 

22. 

64:XSl—Aiis.    568 

CASE    III. 
Art.  153   .    To  mult ipli/ one  fraction  ly  another. 

1.  What  is  the  product  of  |  by  |  ? 

Solution. — To    multiply    ^    by    |,    is   to         operation. 
take   I  of  4   (Art.  152),  and  this  is  equal     ^  X  f  =  |f.  Ans. 
to4X|  =  H(Art.  131). 

Analysis. — Now,  |  of  ^  is  3  times  ^  of  ^;  and  |  of  -|  is  4 
times  i  of  ^;  but  I  of  4  =  3^5  (Art.  130);  hence,  ^  of  ^  is  4 
times  Jp  and  |  of  ^  is  3  times  -^o^^  H-  '^^^^  ^^  multiplying 
the  numerators  together,  and  the  denominators  together. 

*2.  What  is  the  product  of  f  by  |  ?    .     .     Ans.  r^^. 
*3.     The  product  of  J  by  f  ?      .     .     .     .     Ans.  if. 

Rule  for  Case  111.— Multipli/  together  the  numerators  for  a 
new  numerator^  and  the  denominators  for  a  new  denominator. 

Rem. — Multiplying  a  fraction  by  a  fraction,  is  the  same  as 
reducing  a  compound  to  a  simple  fraction. 


4. 
5. 


/g  X  I  =  Ans. 


1 1  V  5 
T2  ^   y 


27 

Ans.  -jS^s.. 


6. 

7. 


8_ 
1  1 

1  1 

I  (> 


Ans.  -5,t. 

I  b  o 


i  =  Ans. 


I  1 
3i5- 


% 


Review.— 152.  Rem.  2.  "Why  may  the  examples  in  this  and  tho 
preceding  case  bo  solved  by  tho  sauio  rule? 

152.  Rem.  3,  When  is  tho  multiplicand  ji/^cr^a*c(/ by  multiplying?  When 
decreased  f    4.  What  oporatiun  is  involved  in  multiplying  by  a  fraction  ? 


COMMON   FRACTIONS.  161 

Note  1, — Reduce  mixed  numbers  to  improper  fractions. 

8.  Multiply  2]  by  31. 

Sol.— 2i  =|,  and  3^  =  J  ;  then  |  X  J  =  6_3  ==  7?   J^^s. 


9.  8|Xf.  Ans.  3|. 

10.  -9^X17/-.  Ans.  15 p\. 


11.  2iX    2A.       Ans.  61. 

12.  leixiei  Ans.  2721. 


• 

Ans 

■  n- 

• 

Ans 

• 

Ans. 

781. 

A 

ns.  417^4. 

Ans. 

17?. 

— 

Ans 

1 

= 

Ans 

1. 

= 

Ans 

5 

= 

Ans. 

24. 

Note  2. — After  indicating  the  operation,  employ  Cancellation, 
when  possible,  as  shown  in  Art.  61. 

13.  Multiply  i    of  4    by   ^   of  5. 

14.  Multiply   f   of  ^  by   1    of  3^. 

15.  31ultiply  I   of  91  by   |    of  17. 

16.  Multiply  -,%  of  7i  by  ji  of  86-,3-. 

17.  Multiply  3,  21,  f ,  and  |  of  5  together, 

18-         iX^3_x8x|x|Xf 

19.  IX  I  X4X|X|Xi  of  6 

20.  ^XlXl|Xiof|X|X|of20       = 

21.  2AX6|X3iX/3  of2Xf 
Jg^""  For  additional  problems,  see  Ray's  Test  Examples. 

Art.   153   .  MISCELLANEOUS    EXAMPLES. 
What  will  be  the  cost 

1.  Of  2J  lb.  of  meat,  at  11  cts.  a  lb.  ?        Ans.  24  ets. 

2.  Of  3    yd.  linen,  at  $f  a  yd.  ?  of  5  yd.?  of  7  yd.  ? 

of  61  yd.  ?  5|  yd.  ?  Ans.  to  last,  ^^. 

3.  Of  3|  lb.  of  rice,  at  4|  cts.  a  lb.  ?         Ans.  16  cts. 

4.  Of  3  J  tuns  of  iron,  at  818|  per  T.  ?         Ans.  $60. 

5.  Of  I5  yd.  of  muslin,  at  $^\  per  yd.  ?         Ans.  $i. 

6.  Of  24  lb.  of  tea,  at  ^  per  lb.  ?  ^ws.  $2. 

7.  Of  55-  cords  of  wood,  at  U  J  per  C.  ?        Ans.  $6f . 

Eeview. — 152.  How  may  a  whole  be  multiplied  by  a  mixed  number? 
153^.  How  multiply  one  fraction  by  another,  Rule  ?  Note  1.  What  is  re- 
quired when  mixed  numbers  occur  ?    2.  When  employ  Cancellation  ? 

3d  Bk.  11 


162  RAY'S   PRACTICAL   ARITHMETIC. 

8.  At  the  rate  of  5^  miles  an  hour,  how  far  will  a  man 
travel  in  7|  hours?  Ans.  42|  mi. 

9.  I  own  I  of  a  steamboat,  and  sell  |  of  my  share  : 
what  part  of  the  boat  do  I  sell  ?  Ans.  |. 

10.  At  $6|   per  yard,  what  cost  |   of  a  piece  of  cloth 
containing  5^  yards?  Ans.  $8]. 

11.  f  of  §  of  16A,  X  I  of  I  of  15,=what?     Ans.  343. 
S^^  For  additional  problems,  see  Ray's  Test  Examples. 

Art.  154.  DIVISION    OF    FRACTIONS. 
Case  I.   To  divide  a  fraction  hy  a  whole  number. 

The  object  in  dividing  a  fraction  by  a  whole  number,  is  to 
separate  the  fraction  'nto  a  given  number  of  equal  parts,  and 
find  the  value  of  one  of  the  parts ;  or,  to  find  what  part  a 
fraction  is  of  a  whole  number. 

1.  If  3  yards  of  ribbon  cost  ^Ij,  what  cost  1  yard? 
Analysis. — If  3  yards  cost  ^  of  ^„„r.. .„,..„ 

•^  T  OPERATION. 

a  dollar,  1  yard,  being  i  of  3  yards,         r-.  q\p 

will  cost  1  of  ^  =  3  of  a  dollar.  _  -^  3  =    /^  =  ~  Ans. 

Or,   since  a  fraction   is  divided  j^ 

by    multiplying    its    denominator  Or, 

(Art.  133),  multiply  7  by  3,  and  the      ^  n 

result,  2^,  reduced,  is  ^,  as  by  the      -  H-  3  =  ~-—z=i^^  =  j. 
first  method.  ] 

Here,  |  is  divided  into  3  equal  parts,  and  the  value  of  each  part 
is  |;  thus,  ^  =  ^  +  ^-(-^;  the  number  of  parts  corresponds  to  the 
divisor,  and  the  value  of  each,  to  the  quotient.  ^ 

2.  At  2  dollars  a  yard,  what  part  of  a  yard  of  cloth 
can  be  bought  for  ^  a  dollar  ? 

Analysis. — Had    it    been    required    to  operation. 

find  how  many  yards,  at  $2  a  yard,  could     i  i  -i 

be  bought  for  $f),  tlie  $G  should  be  divided     o  "^~'^X*^^^4 
by  $2;  and,  to  find  (he  part  of  a  yard  that 
$^.  will  pay  for,  divide $»  by  $2:  to  divide^  ^*«»-  :}  Jf^- 

by  2,  multiply  thedenoniiuator(  Art.  133);  the  quotient  is  one-fourth. 


COMMON   FRACTIONS.  103 

*3.  If  4  yards  of  muslin  cost  |  of  a  dollar,  what  will 
1  yard  cost?  Ans.  $'^. 

*4.  If  1  orange  cost  3  cents,  wliat  part  of  an  orange 
can  be  purchased  for  A  a  cent  ?  Ans.  J. 

Rule  for  Case  I. — Divide  the  numerator  by  the  whole  num- 
ber^ when  it  can  be  done  without  a  remainder ;  write  the  quotient 
over  the  denominator.  Otherwise,  Multiply  the  denominator  by 
the  whole  number^  and  write  the  product  under  the  numerator. 

5.  If  4  yards  of  cloth  cost  ||  of  a  dollar,  what  will 
1  yard  cost?  Ans.  % ^q. 

6.  If  a  man  travel  j^^^  of  a  mile  in  3  hours,  how  far 
does  he  travel  in  1  hour  ?  Ans.  -^^  mi. 

7.  If  5  yards  of  tape  cost  ^q  of  a  dollar,  what  will 
1  yard  cost?  Ans.  j^q. 

8.  If  7  pounds  of  coffee  cost  ^4  of  a  dollar,  what 
will  1  pound  cost  ?  Ans.  $tVv. 

9.  At  4  dollars  a  yard  for  cloth,  what  part  of  a  yard 
will  i|  of  a  dollar  buy?  Ans.  o^yd. 

10.  At  5    dollars   a   tun,    what  part    of   a   tun   of  hay 
will  i  a  dollar  purchase  ?  Ans.  j'^  T. 

11.  At  6  dollars  a  barrel,  what  part  of  a  barrel  of  flour 
will  S2|  pay  for  ? 

T>     •,  -1  ,         ^  .  OPERATION. 

Keduce  a  mixed  number  to  an  improper  , 

fraction,  and  it  may  be  divided  by  a  whole  o         5 


number,  the  same  as  a  proper  fraction. 


Jv2-^6  =  i  Ans. 


12.  If  5  bushels    of  wheat   cost   3|   dollars,  what   cost 
1  bushel  ?  Ans.  |  of  a  dollar. 

I 

13.  If  7   ounces   of  opium    cost   8|   dollars,  what  cost 

1  ounce?  Ans.  -5  =$14. 

14.  If  J I  be  divided  into  9  equal  parts,  what  will  each 
part  be?  Ans.  -=-2 

P  L!_ 

Review. — 154.  What  is  the  object  in  dividing  a  fraction  by  a  -whole 
jl    number  ?     In  Ex.  1,  what  does  the  divisor  show  ?     The  quotient  ? 

154.  In  Ex.  2,  what  does  the  quotient  show  ?     How  divide  a  fraction  by 
a  whole  number,  Rule  for  Case  I  ? 


164 


RAY'S   PRACTICAL    ARITHMETIC. 


15. 

4|    :     8  -  Ans.    |. 

18. 

33- 

-   7—  Ans.    Ij 

16. 

1  H-    5  —  Ans.  ^%. 

19. 

47f- 

-15—  Ans.  3 A 

17. 

124    :   11  -  Ans.  1]. 

20. 

130^- 

-18=  Ans.7£^ 

CASE   II. 
Art.  155.    To  divide  a  whoU  number  hy  a  fraction. 

Dividing  a  whole  number  by  a  fraction,  is  finding  liow 
many  times  the  fraction  is  contained  in  the  whole  number. 

1.  At  I  of  a  cent  for  1  lemon,  how  many  can  be  bought 
for  4  cents  ? 


OPERATION. 

4 
3 


Solution. — In    4    cents    there   are    12 

thirds  of  1   cent   (Art.  141).     If  1  lemon 

costs  2  thirds  of  a  cent,  there  will  be  as    mi,-   i     om  o   io,-   i 

,                 r»    ,  •    ,                                        Ihirds  J)i.i   thirds, 
many  lemons  as  2  thirds  are  contained  times  — 

in  12  thirds;  that  is,  6.     Ans.  6  lemons.  Ans.  6 

Or  thus:  ^  is  contained  in  4  as  many  times  as  there  are  thirds 
in  4,  that  is,  12  times;  and  2  thirds  in  4,  one  half  as  many  times 
as  1  third;  12-1-2=6  times. 

In  this  operation,  the  whole  number  is  reduced  to  the 
same  name — the  same  parts  of  a  unit — as  the  divisor, 
that  the  divisor  and  dividend  may  be  of  the  same  de- 
nomination.    Art.  41,  Rem. 

The  whole  number  is  multiplied  by  the  denominator  of 
the  fraction,  and  the  product  divided  by  the  numerator. 

*2.  At  },  a  cent  each,  how  many  apples  can  be  bought 
for  3  cents^?  Ans.  6  apples. 

*3.  At  I  of  a  dollar  per  yard,  how  many  yards  of 
cloth  can  you  buy  for  G  dollars?  Ans.  8  yards. 

Kule  for  Case  II. — Multiply  the  whole  nvmher  hy  the  denom- 
inator of  the  fraction,  and  divide  the  product  by  the  numerator. 

Rkvirw, — lr>5.  What  is  dividing  a  wholo  number  by  a  fraction? 
IIow  many  times  is  one-third  contained  in  1  ?  In  2?  In  8  ?  In  4? 
How  many  times  two-thirds  in  1  ?     In  2?     In  3?     In  4? 


COMMON   FRACTIONS.  165 


4. 

4- 

=-  1  =  Am.    10. 

7. 

6   :     3   —  ^«s.    14. 

0. 

16- 

-  1   —  Am.  2\\. 

8. 

13  :    1  =  Ans.  21|. 

6. 

8- 

-  j%  =  Ans.    60. 

9. 

21   :    ,\=  Ans.    33. 

10.  In  one  rod  there  are  5^  yards  :  how  many  rods  are 
there  in  22  yards  ? 

Reduce  a  mixed  number  to      operation       5  ^  =  U 
an    improper    fraction,    and    a 

9*^  V  9         4-4. 

whole  number  may  be  divided      -'-'  '^  --L  — ^^"=4  rods   y^?7S 


by  it,  as  by  a  proper  fraction. 


11        11 


I 


11.  At  2|  dollars  for  1  yard  of  cloth,  how  many  yards 
can  be  bought  for  $6  ?  Ans.  2J  yd. 

12.  At  3|  cents  a  lb.,  how  many  pounds  of  rice  caji 
be  bought  for  30  cts.  ?  Ans.  8  lb. 

13.  How  many  times  4|  in  50?  Ans.  ll/y. 

14.  Divide  56  by  54.  Ans.  10|-. 

CASE   III. 
Art.  156.    To  divide  a  fraction  ht/ a  fraction. 

The  object  in  dividing  a  fraction  by  a  fraction,  is  to 
find  how  many  times  the  divisor  is  contained  in  the  divi- 
dend ;  or,  what  part  the  divisor  is  of  the  dividend. 

1.  At  f^  of  a  dollar  per  yard,  how  many  yards  of 
muslin  can  be  bought  for  ^f^  ? 

Suggestion. —  Find    how    often    2    tenths  opebation. 

are  contained  in  9  tenths^  as  you  would  find      Tenths  2)9  tenths. 

how  often  2  cents  are   contained  in  9  cents ;  .         ^T" 

Ans    4— 
that  is,  by  dividing  9  by  2:  9-^2  =  4 J.  '     ^' 

Hence,  when  two  fractions  have  a  common  denominator^  obtain 
their  quotient  by  dividing  the  numerator  of  the  dividend  by 
the  numerator  of  the  divisor. 

2.  How  many  times  |  in  |  ?  operatton. 

We  can  not  find  how  often  2  ^  v  4.  ^^  T%  5   d  V  ^  ^^  T%  * 

inches  are  contained  in  3  feet, 

without  expressing  the  divisor        -9_  _^_?^  =  9-f-8  =  1^  Ans. 
and   dividend  in  the  same   de- 
nomination,   inches;    so,    also,    to    find    how    often    2    thirds    are 


166 


RAY'S    PRACTICAL    ARITHMETIC. 


contained  in  3  fourths,  reduce  them  to  the  same  denomination^ 
twelfths:  I  are  8  twelfths,  and  |  are  9  twelfths;  and  8  twelfths  are 
contained  in  9  twelftfts,  9-i-8  =  IL 

No  use  is  made  of  the  common  denominator,  the  numerator  of  the 
divisor  being  multiplied  by  the  denominator  of  the  dividend,  and 
the  numerator  of  the  dividend  by  the  denominator  of  the  divisor. 

This  is  easily  performed  by  inverting  the  terms  of  the  divisor, 
then  proceeding  as  in  Multiplication  of  Fractions,  Art.  15-3. 

Thus,  I  --§  =  I  X  I  =  i  =  1^  Ans. 


I    be    divided    by    2,    the    quotient, 


Another    Solution. — If 
3 
(Art.  133),  is  J    f,;   but,   since  2  is  3  times  |,  the  divisor  used 

is  3  times  the  given  divisor;  hence,  multiply  this  quotient  by  3  to 
obtain  the  true  quotient:  this  gives  |X|  =  f=lg  -4^*' 

*3.  At  I  of  a  dollar  each,   how  many  knives  can  you 
buy  for  J  of  a  dollar  ?  Ans.  2. 

*4.  At   1   of  a   dollar  per   yard,   how  many  yards  of 
ribbon  can  be  purchased  for  |  of  a  dollar  ? 


Ans.  3|. 


Rule  for  Case  III. — Invert  the  divisor ;  multiply  the  nu- 
merators together  for  a  new  numerator,  and  the  denominators 
for  a  new  denominator. 

Note. — Reduce  compound  to  simple  fractions,  and  mixed  num- 
bers to  improper  fractions,  before  commencing  the  operation. 


I 


FIND   THE   QUOTIENT   OF 


5. 

1  H-  i   ;=  Ans.  3. 

10. 

2A- 

_    1 

T(T 

—  Ans.  40. 

6. 

y,    :    }   =  Ans.  2. 

11. 

U- 

-n 

=  Ans.  33. 

7. 

1  -!-  -^  —   Ans.    ^. 

12. 

43- 

-H 

=  Ans.  If. 

8. 

u   '    3          An^.  1^. 

13. 

2.1- 

-n 

—  Ans.  -p3^. 

9. 

A-f-J    —   Ans.  lA. 

U. 

1 

3 

-A 

—  Ans.  25. 

15. 

Divide  |  of  §  by  f  of  |. 

• 

•        • 

Ans.  ijf. 

16. 

Divide  J  of  5^1  by  |  of  17A 

•           • 

•         • 

Ans.  3V^. 

Review. — 155.  How  do  you  divide  a  whole  number   by  a   fraction? 
156.  What  is  tho  object  in  dividing  a  fraction  by  a  fraction  ? 


COMMON    FRACTIONS.  167 

Art.  157.  The  rules  in  the  three  preceding  Articles 
may  be  embraced  in  this 

General  Rule  for  Division  of  Fractions. — Express  the 
divisor  and  dividend  in  the  form,  of  a  fraction  ;  invert  the  di- 
visor ;  cancel  all  the  factors  common  to  both  terms ;  then  multiply 
together  the  numbers  remaining  in  the  numerator  for  a  new 
numerator,  those  in  the  denominator  for  a  new  denominator. 

1.  Divide  A  of  -^  by  1  of  4  '        operatiox. 

In  this  operation,  cancel  the  factors  2  ^  ^ 


and  3  on  both  sides — above  and  below  the      ^  X  o  X  p  X  ■4 —  l-i 
line — then  multiply  together  the  factors  remaining  on  each  side. 


14  -^-  21  =  Ans. 


1  5 


9 

4.     8 


4  5" 


3.     8^-^35  =  Ans.  U. 


4  5 

If  =  Ans.  S!, 
8.  Divide  fg  of  |  of  12/o  by  i  of  Sf     .     .     Ans.  | 


5.     14   --8|  =  Ans.  U. 

4  o 

U.  3  2     •     48    —    -^'ti>.    1  0' 

7.     12|--4i  =  Ans.  31. 


REDUCTION   OF   COMPLEX   TO   SIMPLE    FRACTIONS. 

Art.  158^.  A  complex  fraction  is  the  expression  of  an 

unexecuted  division  (Art.  125),  in  which  the  divisor  or 

dividend,  or  both,  are  fractions. 
11 
Thus,   ^  indicates  that  1|  is  to  be  divided  by  2J. 

Hence,  to  reduce  a  complex  to  a  simple  fraction, 

Regard  the  numerator  as  a  dividend,  the  denominator  as  a 
di'msor,  and  proceed  as  in  Division  of  Fractions,  Art.  157. 

11 
1.  Reduce  «!   to  a  simple  fraction. 


OPERATIOX. 


1  1  =5  'i 

4  4    C.5_::_7  —  5y3 — 15        A^o 


In  this  operation,  after  reducing  the  mixed  numbers  to  im- 
proper fractions,  the  numerator  of  the  result  is  the  product  of  the 

Eeview. — 156.  When  two  fractions  have  a  common  denominator,  how 
obtain  their  quotient  ?  How  find  how  often  2  inches  are  contained  in  3  feet  ? 
How  often  two-thirds  are  contained  in  three-fourths  ?  How  divide  a  frac- 
tion by  a  fraction  ?    157.  "What  is  the  General  Eule  ? 


168 


RAY'S   PRACTICAL    ARITHMETIC. 


extreme  terms,  5  and  3  ;  the  denominator,  the  product  of  the  mean 
terms,  4  and  7.     Hence, 

To  Reduce  a  Complex  to  a  Simple  Fraction, 

Reduce  mixed  numbers  to  improper  fractions^  then  multiply 
together  the  extreme  terms  for  a  numerator^  and  the  mean  terms 
for  a  denominator. 

Reduce  these  complex  to  simple  fractions  : 

0.    -^  —  An&  J  4 


f5 

2.    ^=zAns    3  0 

1_1         ^"'^'    IT 
5 

2 

^-    -|-  =Ans.  ft. 
5  *  ^ 


4         — 

Q~>         ./i.nS,    "ry. 


31 
0 .      —  A  tiQ     1  T  5 

^7 


3| 


2 
3" 


Complex  fractions  may  be  multiplied  together,  or,  one  divided 
by  another,  by  first  reducing  each  to  a  simple  fraction.  Indicate 
the  operation,  and  cancel. 

8.  Multiply  9i  by  1^ ^«s-  A- 

9.  Multiply  ^  of-H  by  y ^«s-  If- 


1  5 


(5 


1^  51 

10.  Divide  ^  by^, 


^7W.  11. 


Art.  158^  miscellaneous    examples. 

1.  At  .^  a  dollar  per  yd.,  how  many  yards  of  silk  can  be 
bought  for  ^3i  ?  Ans.  ^  yd. 

2.  At  I  of  a  dollar  per  pound,  how  many  pounds  of  tea 
can  be  purchased  for  %2,-^^  ?  Ans.  3|lb. 

3.  Find  the  quotient  of  A  divided  by  2 ;    by  A  ;    by  -J  ; 
by  I  ;  by  J  ;  by  -j-J^^.  '  Ans.  to  last,  500. 

4.  At  3J   dollars  per  yard  for  cloth,  how  many  yards 
can  be  purchased  with  $42 1  ?  Ans.  ll|yd. 

5.  By  what  must  |  be  multiplied,  that  the  product  may 
be  10?  J/is.  26|. 

Review. — ISS'*.   "What  is  a  complex  fraction  ?    How  reduced  to  a 
simple  fraction  ? 


1 


COMMON   FRACTIONS.  169 


a' 


6.  Divide  3^  by  |  of  li Ans.  5' 

7.  Divide  j\  of  271  by  j%  of  211.       .     .  Ans.  Iff. 

14        2i 

8.  Divide  ir  by  «f -^^*-  ^3  2* 

1  "^6 

Art.  159.  examples  in  u.  s.  money. 

1.  Add  $16,061;  §9.12.1;  $5,433;  §2.81i 

Ans.  $33.4S| 

2.  I  paid  for  books  89.121;    paper  $4.43|  ;    a   slate 
.37^^;  quills  ^1.621:  what  did  I  pay  ?    .4«s.  $15,561 

3.  Having   $50.25,    I    paid   a    bill   of  S27.18| :    how 
much  had  I  left?  Ans.  $23,061 

4.  From  $32,311  take  $15.12^  Ans.  $17.18| 

5.  From  $5,811  take  $1.18|  Ans.  $4.62i 

Find  the  cost  of 

6.  9  yd.  of  muslin  at  12Acts.  a  yd.  Ans.  $1.12^ 

7.  21  lb.  of  sugar  at  6 lets,  a  lb.  Ans.  $1,311 

8.  15yd.  of  cloth  at  $3.18|  per  yd.  Ans.  $47,811 

9.  51  yd.  of  linen  at  $0 .  62 A  per  yd.  Ans.  $3 .  43| 

10.  12Ayd.of  ribbonat  18|cts.peryd.  ^ns.  $2.34| 

11.  131  yd.  of  calico  at  16|cts.  per  yd.     Ans.  $2.25 

12.  lOiyd.  of  cloth  at  $3.37^  a  yd.     Ans.  $34.59| 

13.  17f  dozen  books  at  $3.75  per  doz.  Ans.  $66.25 

14.  At  18|cts.   per  yd.,  how  many  yards  of  muslin  can 
be  purchased  for  $2.25  ?  ^?is.  12yd. 

15.  At  37^  cents  per  bu.,  how  many  bushels  of  barley 
can  you  buy'for  $5,811  ?  jlns.  151  bu. 

16.  If  five  yards  of  cloth  cost  $11.56i,  what  cost  one 
yard?  Ans.  $2,311 

17.  Seven  men  share  $31,061  equally:  what  is  the  share 
of  each  man?  ^/is.  $4.43| 


170 


RAY  S  PRACTICAL   ARITHMETIC. 


EXAMPLES    IN     LONG    MEASURE. 

18.  Reduce  5  mi.  to  inches.        .     .     Ans.  316800  in. 

19.  2  mi.  2  rd.  2  ft.  to  feet.        .     .     Ans.    10595  ft. 

20.  20  yd.  to  rods. 

OPERATION. 

5^  yards  =  11  halves,  20  yards  =  40  halves. 
11)40 

3  rd.  7  half  yd.  left,  =  3^  yd.       A  ns.  3  rd.  3^  yd. 

Suggestion. — In  reducing  numbers  from  a  lower  to  a  higher 
denomination,  when  the  divisor  is  a  fractional  number  (Art.  155), 
reduce  both  divisor  and  dividend  to  like  parts  of  a  unit. 

The  remainder  being  of  the  same  denomination  as  the  dividend, 
(Art.  38),  will  be  a  fraction,  which  reduce  to  a  whole  or  mixed 
number.     Here,  the  remainder,  7  half  yd.,  reduced,  makes  3^  yd. 

21.  Eeduce  15875  ft.  to  miles.      Ans.  3  mi.  2  rd.  2  ft. 

22.  142634  in.  to  miles.    A  hs.  2  mi.  2  fur.  2  yd.  2  in. 

23.  How  many  steps,  of  2  ft.  8  in.  each,  will  a  man  take 


in  walking;  2  miles? 


Ans.  3960. 


24.  How   many  revolutions  will  a  wheel,  of  9  ft.  2  in. 
circumference,  make,  in  running  65  mi.?       Ans.  37440. 


EXAMPLES   IN   SQUARE   MEASURE. 

25.  Reduce  1  A.  3R.  16  P.  25  sq.  yd.  to  square  yards. 

Ans.  8979  sq.  yd. 

26.  7506  sq.  yd.  to  A.    Ans.  1  A.  2  R.  8  P.  4  sq.  yd. 

27.  5  chains  15  links,  to  in.  Ans.  4078i  in. 

28.  How  many  acres  in  a  field  40'.  rd.  Ions,  and  32  rd. 
wide?  "    Ans.SA.161\ 

EXAMPLES    IN   TIME   MEASURE. 
In  these  exan)ples,  the  year  is  supposed  to  be  3654  days. 

29.  Reduce  4  years  to  hours.  .     .     .     ^7?s.  35064  hr. 

30.  914092  hr.  to  cen.     Ans.  leen.  4yr.  101  da.  4hr. 

31.  In  what  time  will  a  body  move  from  the  earth  to 
the  moon,  nt  the  rate  of  31  miles  per  day,  the  distance 
being  238545  miles? 


Ans.  21  yr.  244Ua. 


I 


COMMON   FRACTIONS.  171 

REDUCTION    OF    FRACTIONAL    COMPOUND    NUMBERS. 

CASE   I. 

Art.  160.    To  reduce  a  fraction  of  a  higher  denomina- 
tion^ to  a  fraction  of  a  lower. 

1.  Reduce  -^^  of  a  peck  to  the  fraction  of  a  pint. 

Solution. — To  reduce  pecks  to  pints,  operation. 

we    multiply  by   8  to  reduce    them   to    pk.  qt.  qt.  pt. 

quarts,  then    by  2    to  reduce  them  to      ivfi i'iv9 2. 

^  54  ^®~3J   3^-^  —  3  • 

pints. 

In  like  manner,  multiply  ihe  fraction  or, 

of  a  peck  by  8,  to  reduce  it  to  the  frac-      i   y  Q\>'9 2  r»f     A 

tion  of  a  quart,    then   by  2,  to  reduce    ^"^  3  r  ' 

it  to  the  fraction  of  a  pint. 

Hence,  fractional  numbers  may  be  reduced  from  a  higher  to  a 
lower  denomination,  (Art.  81),  by  multiplying  by  that  number  of 
the  next  lower  order  which  makes  a  unit  of  the  higher. 

*2.  Reduce  ^^  of  a  bu.  to  the  fraction  of  a  qt.    Ans.  |. 

Kule  for  Case  I. — Midtiply  as  in  Reduction  of  Whole 
Numbers,  Art.  81,  according  to  the  rules  for  the  multiplication 
of  fractions. 

Rem. — The  work  in  Cases  I  and  II  may  often  be  shortened  by 
Cancellation. 

3.  Reduce  Tj'g  lb.  Av.  to  the  fraction  of  an  oz.  Ans.    4. 

4.  j'g    of  a  lb.  Troy,  to  the  fraction  of  an  oz.  Ans.    |. 

5.  37'g    of  a  yd.  to  the  fraction  of  a  na.         Ans.    i. 

6.  jT^go  of  an  A.  to  the  fraction  of  a  P.         Ans.    |. 

7.  -gl^  of  a  dollar  to  the  fraction  of  a  ct.      Ans.    |. 

8.  ^554  of  a  da.  to  the  fraction  of  a  min.       Ans.  \^. 

9.  3!^  of  a  bu.  to  the  fraction  of  a  pt.         Ans.    |. 

Review. — 160.  How  reduce  pecks  to  pints  ?  How  the  fraction  of  a 
peck  to  the  fraction  of  a  pint?  How  are  fractional  numbers  reduced 
from  a  higher  to  a  lower  denomination  ? 


172  RAYS   PRACTICAL  ARITHMETIC. 

CASE    XL 

Art.  161.  To  reduce  a  fraction  of  a  lower  denomina- 
tion, to  a  fraction  of  a  hie/her. 

1.  Reduce  |  of  a  pint  to  the  fraction  of  a  peck. 

Solution. — To  reduce  pints  to  pecks,  operation. 

we   first  divide   by  2  to  reduce  them  pt.  qt.    qt.  pk. 

to  quarts,    and   then  by  8  to  reduce    ;^_^2=4^  *   -?-i-8  =  J-  : 

them  to  pecks, 

or 
In  like  manner,  divide  the  fraction  ' 

by  2,  to  reduce  it  to  the  fraction  of  a    3  ^ '2 -^  8  ~24  P*'*  '^^^' 

quart,  and  then  by  8,  to  reduce  it  to  the  fraction  of  a  peck. 

Hence,  fractional  numbers  may  be  reduced  from  a  lower  to  a 
higher  denomination  (Art.  81),  by  dividing  the  given  fraction  by 
that  number  of  its  own  denomination  which  makes  a  unit  of  the 
next  higher. 

*2.   Reduce  |  of  a  qt,  to  the  fraction  of  a  bu.     Ans.  Jq. 

Rule  for  Case  II. — Divide,  as  in  Reduction  of  Whole 
Numbers,  Art.  81,  according  to  the  rules  for  the  division  of 
fractions. 

3.  Reduce  5  of  a  na.  to  the  fraction  of  a  yd.     Ans.  -^q. 

4.  f  of  a  gr.  T.  to  the  fraction  of  a  lb.  Ans.  ^g^oo* 

5.  I  of  a  9  to  the  fraction  of  a  tb.  Ans.  g|g. 

6.  I  of  a  pt,  to  the  fraction  of  a  bu.  Ans.  -j^^. 

7.  ^  of  an  oz.  to  the  fraction  of  a  cwt.  Ans.  ^ggg- 

8.  I  of  an  in.  to  the  fraction  of  an  E.En.     Ans.  ^'^. 

9.  I  of  a  min.  to  the  fraction  of  a  da.  Ajis.  yg^j^. 
10.  I  of  a  dr.  to  the  fraction  of  a  qr.  Ans.  ^^q^- 

Review.— 100.  What  is  Case  1?  What  is  the  rule  for  Cjuio  1? 
161.  How  reduce  pints  to  pecks?  How  the  frjvction  of  a  pint  to  the  frac- 
tion of  a  peck  ?  How  are  fractional  numbers  reduced  from  a  lower  to  a 
higher  denomination  ?     What  is  Case  2?     The  Rule  for  Case  2? 


COMMON    FRACTIONS.  173 

CASE    III. 

Art,  162.    To  find  the  value  of  a  fraction  in  integers 
(whole  numbers)  of  a  lower  denomination. 

1.  Find  the  value  of  |  of  a  day  in  integers. 

Solution, — S  of  a  da,,  are  |  of  24  operation. 

hr. ;   and   |  of  24  hr.  are  found  by  24  lir,             6  0  min. 

multiplying  by  2  and  dividing  by  5  ^                      ^ 

(Art.  152).     This  gives  9|  hr.  r~^         5)180 

Again,  i  of  an  hour  are  the  same        

as  I  of  60  minutes,  which  are  36  min.  9|  hr.           3  6  min. 

Hence,  4  of  a  day  =9hr.  36min.  Ans.  9  hr.     36  min. 


*2.  Value  of  4  of  a  mi.  in  intejrers.  Ans.  6  fur.  16  rd. 

Kule  for  Case  III. —  Take  the  number  of  units  of  the  next 
lower  denomination  which  forms  a  unit  of  the  denomination  of 
the  fraction;  multiply  it  by  the  numerator,  and  divide  the 
product  by  the  denominator. 

If  this  division  produce  a  fraction,  find  its  value  in  the  same 
manner,  and  so  on :  the  several  quotients  will  be  the  answer. 

3.  What  is  the  value  of  ^  of  a  dollar?       Ans.  60cts. 


4.  Of  I  of  a  mile  ?      . 

5.  Of  i  of  a  lb.  Troy  ? 


6.  Of  I  of  a  lb.  Av,  ?       , 

7.  Of  I  of  an  acre  ? 

8.  Of  ^  of  a  T.  of  wine  ? 


Ans.  3  fur.  8rd. 

.     .  Ans.  9oz.  12pwt. 

.     .     Ans.  9oz.  2 1  dr. 

.     .       J«s.  2R.  20P. 

Ans.  3hhd.  31  gal.  2qt. 


I 


CASE    IV. 

Art.  163,  To  reduce  a  quantity  having  one  or  more 
denominations^  to  the  fraction  of  another  quantity  composed 
of  one  or  more  denominations. 

1.  Reduce  2  feet  3  inches  to  the  fraction  of  a  yard  : 
that  is,  2ft.  3 in,  is  what  part  of  1  yard? 

■      Solution. — 2  ft,    3  in.  =  27    inches:     1  yd.  =  36    inches;    and 
since  1  inch  is  ^',  of  36  inches,  27  inches  are  i|=  §.      Ans.  3  yd. 

3b  '  36         4  4  *' 


1 


174  RAY'S   PRACTICAL    ARITHMETIC. 

2.  Find  what  part  2ft.  6 in.  is  of  6ft.  Sin. 

c  T.     1       •  1.    .1-  i-i-  OPERATION. 

Solution. — Reducing   both   quantities 

to  the  same  denomination,  the  first  is  30,  ^  ^*-  ^  ^°-  ~  ^^  J°' 

and  the  second  80  in.     Since  1  in.  is  g'^j  "  ^^-  ^  ^"-  =  ^^  ^°- 

of  80  in.,  30  in.  will  be  |0^|.  Aiis.  |g  =  |. 

*3.  Reduce  2pk.  4qt.  to  the  fraction  of  a  bu.  Ans.  |. 
*4:.  What  part  is  2  yd.  Iqr.    of  8  yd.  3qr.?    Ans.  -^^. 

Rule  for  Case  IV. — Reduce  both  quantities  to  the  lowest 
denomination  in  either ;  the  less  will  be  the  numerator,  and  the 
greater  the  denominator  of  the  required  fraction,  which  reduce 
to  its  lowest  terms. 

5.  Reduce  13  hr.  30  min.   to  thefrac.  of  ada.    Ans.  ^^.. 

6.  3fur.  25  rd.  to  the  fraction  of  a  mile.      Ans.  §|. 

7.  2ft.  Sin.  to  the  fraction  of  a  yard.  Ans.  |. 
S.  What  part  is  96  pages  of  432  pages  ?  Ans.  |. 
9.       15 mi.  3 fur.  3rd.    of  35 mi.  7 fur.  7rd.?    Ans.  |. 

10.  A  man  has  a  farm  of  168  A.  28  P. :  if  he  sell  37  A. 
2R.  14P.,  what  part  will  he  sell?  Ans.  ^y^. 

Notb:. — If  one  or  both  quantities  contain  a  fraction,  reduce  them 
to  a  common  denominator,  and  compare  their  numerators. 

11.  What  part  is  7oz.  IJdr.  of  1  lb.  Av.?        Ans.    |. 

12.  2qt.  IJpt.  of  Ibu.  Iqt.  l?pt.?  Ans.  J563. 

13.  1yd.  1ft.  1/yin.  of  3yd.  2ft.  8f  in.?  Ans.  ifAf. 

ADDITION    AND    SUBTRACTION 
OF    FRACTIONAL    COMPOUND    NUMBERS. 

Art.  164.   1.  Add  |  of  a  yard,  to  |  of  a  foot. 

Solution.— Find  the  value  of  each  of  the         operation. 

quantities   in   integers  (Art.  1G2),   and   then  3     j   9       o 

obtain  their  sum  by  the  rule  for  addition  in  4  •/    '       ^ 

Art.    101.     If   their   difference    be    required,  (J  ^^'    ~~1_____ 
subtract  according  to  the  Rule  in  Art.  102.  Ans.  3       1 


COMMON   FRACTIONS.  175 

Rule  for  Addition  or  SubtTSLCtion.— First,  Jind  the  value 
of  each  fraction  in  integers  (Art.  162)  : 

Then  add  or  subtract,  as  may  he  required,  according  to  the 
rules  for  the  addition  and  subtraction  of  compound  numbers. 

Note. — If  fractions  occur  in  the  lowest,  denomination,  add  or 
subtract  by  the  rules  for  addition  and  subtraction  of  fractions. 

2.  Add  f  of  a  da.  to  |  of  an  hr.      Ans.  16 hr.  45min. 

3.  Add  together  ^  wk.  |da.  Jhr.      Ans.  2 da.  15min. 

4.  |wk.  I  da.  |lir.  |  min.         Ans.  5  da.  6hr.  40  sec. 

5.  I^Jgal.  of -v^ine,  J^hlid.         ^7?s.  6gal.  Ipt.  l|gi. 

6.  How  much  land  in  2^R.,  |  A.,  and  3R.  28|P.? 

Ans.  2  A.  IR.  9-JiP. 

7.  I  of  a  da.,  less  j^g  of  an  hr.,  equals  what? 

Ans.  18 hr.  36 min.  40  sec. 

8.  I  of  $1,  less  I  of  a  dime?  Ans.  55cts. 

9.  I  of  an  oz.,  less  |  of  a  pwt.  ?     Ans.  llpwt.  3gr. 

10.  fmi.,  less  j''^fur.  ?       Ans.  Ifur.  5rd.  10ft.  lOin. 

11.  ^da.,  less  |hr.  ?  Ans.  19 hr.  54 min.  IT^sec. 

12.  |E.En.,  less  §  yard?  Ans.  3qr.  j|  na. 

Art.  165.   PROMISCUOUS    EXAMPLES. 

1.  Reduce  |||f  f  to  its  lowest  terms.  .     .     Ans.  ji. 

2.  Find  the  sum  of  |  and  §  ;  of  |  and  ^  ;  of  -f^  and  1 1 ; 
of  2.^,  3|,  f5,  and  ^^.  Ans.  to  last,  6if .  "^ 

3.  What  is  the  difference  between  |  and  f  ?  ^  and  j\  ? 
34  and  14?  3§  and  \  of  3-]?  Ans.  to  last,  2^^. 

4.  Add  f  of  fo  to  I  of  /^ Ans.  f  |. 

Review. — 162.  Two-fifths  of  a  day,  are  two-fifths  of  what?  How  find 
two-fifths  of  24  hours  ?     What  is  Case  3  ?     What  is  the  Rule  ? 

163.  What  is  Case  4?  What  the  Rule?  Note.  If  one  or  both  quan- 
tities contain  a  fraction,  how  proceed?  164.  What  is  the  rule  for  the 
addition  or  subtraction  of  fractional  compound  numbers  ? 


176  RAY'S    PRACTICAL   ARITHMETIC. 

5.  To   the   quotient    of   1|-t-2.^,    add    the    quotient 
of  51^3^.  '  Ans.m. 

6.  What  number  divided  by  |  will  give  10  for  a  quo- 
tient? Ans.  6. 

7.  What  number  multiplied  by  |  will  give  10  for  a 
product?  Ans.  16|. 

8.  What  number  is  that,  from  which  if  you  take  f  of 
itself,  the  remainder  will  be  16?  Ans.  28. 

9.  What  number   is  that,  to  which  if  you  add    ^  of 
itself,  the  sum  will  be  20  ?  Ans.  14.        'jk 

10.  A  boat  is  worth  S900 ;  a  merchant  owns  |  of  it, 
and  sells  *  of  his  share :  what  part  has  he  left,  and  what 
is  it  worth  ?  Am.  j%  left,  worth  $375. 

11.  I  own  ^^  of  a  ship,  and  sell  ^  of  my  share  for 
$1944i  :  what  is  the  whole  ship  worth?     Ans.  $10000. 

12.  What  part  of  3  cents,  is  |  of  2  cents?        A7is.  |. 

13.  What  part  of  3G8,  is  176  ?  Ans.  ^ . 

14.  What  number,  +  ^-|-yy+-fy-p,  =  5|  ?     Ans.  4gg|. 

15.  What  number,  +  /g  of  j%  of  4-p»^,  =  l  ?      Ans.  -^%. 

16.  From  the  quotient  of  §^-f,  subtract  the  quotient 

01   g-^jy-  Ans.  -rf:\Q- 

17.  If  T  walk  2044  rods  in  -j^^  of  an  hour,  at  that  rate 
how  far  will  I  walk  in  1  j4hr.  ?  Ans.  8468 rd. 

18.  What  part  of  1]  feet,  is  3]  inches?  Ans.  |. 

19.  Two  men  bought  a  bl.  of  flour  ;  one  paid  S31,  and 
the  other  $3J  :  what  part  of  it  should  each  have  ? 

Ans.  One  y^y^g,  the  other  fj^^.         | 

20.  A  has  $2400  ;  f  of  his  money,-f  $500,  is  J  of  B's  : 
what  sum  has  B  ?  Ans.  $1600. 


21.  John  Jones  divided  his  estate  among  2  sons  and  3 
daughters,  the  latter  sharing  equally  with  each  other. 
The  younger  son  received  $2200,  which  was  A  of  the 
share  of  the  elder,  his  share  being  ]§  of  the  whole  estate  : 
find  the  share  of  each  daughter.  Ans.  $1356'^. 


I 


t 


XII.    DECIMAL   FRACTIONS. 

Art.  166.  A  common  fraction  (Art.  124)  is  one  whose 
denominator  may  be  any  number ;    as,  2,  3,  4,  &c. 

A  decimal  fraction  is  one  whose  denominator  is  10,  or  a 
number  of  lO's  multiplied  together;  as,  100,  1000,  &c. 

Explanation. — A  common  fraction  arises  from  dividing  a  unit 
into  any  number  of  equal  parts.  A  decimal  fraction  arises  from 
dividing  a  unit  by  10,  or  some  multiple  of  10  by  itself;  and  its 
denominator  is  not  usually  expressed  in  figures. 


Art.  167.  If  a  unit  be  divided  into  10  equal  parts, 
each  part  will  be  1  tenth;    thus,  jL  of  1  =  Jg. 

If  each  tenth  be  divided  into  10  equal  parts,  the  unit 
will  be  divided  into  100  equal  parts,  each  part  being  1 
hundredth,;  thus,  Jq  ^^  t*o=^T^U• 
If  each  hundredth  be  divided  into  10  equal  parts,  the 
unit  will  be  divided  into  1000  parts,  each  part  being 
1  thousandth  of  the  whole  ;  thus,  -^^  of  -j-Jq  =  jx/q  o* 

A  comparison  of  the  fractions  j'^,  j^q,  J(5^oo'  ^^-j  showP 
that  the  value  of  each  is  one-tenth  of  that  which  precede? 
it ;  that  is,  they  decrease  in  value  tenfold. 

Art.  168.  The  orders  of  whole  numbers  decrease  from 
left  to  right  tenfold ;  thus,  1  hundred  is  one-tenth  of  1 
thousand;  1  ten  is  one-tenth  of  1  hundred;  and  1  unit  is 
one-tenth  of  1  ten  :  in  like  manner. 

The  orders  may  be  continued  from  the  place  of  units 
toward  the  right^  by  the  same  law  of  decrease  ;  and, 

The  first  order  on  the  right  of  units  will  then  express 
tenths  of  units,  that  is  tenths;  the  next  order,  tenths  of 
tenths,  or  hundredths ;  the  next,  tenths  of  hundredths, 
or  thousandths  ;  and  so  on  : 

Hence,  if  a  point  (.)  be  placed  to  separate  units  and 
tenths,  j'q  may  be  written  in  the  order  of  tenths ;  thus,  .1, 
as  1  (unit)  is  written  in  the  order  of  units  :  and, 

foi  T%'  T*o'  ^^-i  ^^y  be  written,   .2,    .3,    .4,  &c. 

•  Review. — 166.  What  is  a  Com.  Fraction?    A  Decimal?    167.  What  la 
1  tenth  ?  1  hundredth  ?  1  thousandth  ?    What  the  value  of  each  ? 

3dBk.  12  177 


178  RAY'S   PRACTICAL   ARITHMETIC. 

Also,  jJ  (J,  j§o,  jIq,  &c.,  may  be  written  in  the  order 
of  hundredths^  thus:   .01,   .02,   .03,  &c.  :  here,  .^ 

There  being  no  tenths,  place  a  cipher  in  the  order  of  tenths. 

As  the  orders  decrease  from  left  to  right,  in  the  same  manner 
as  in  whole  numbers,  Decimal  Fractions  may  be  expressed  in 
figures,  without  writing  ilieir  denominators. 

Note. — ^3^,  y2_4_  J2jh5_^  and. 3,  .24,  .205  are  alike  decimal  fractiong.  ^ 

Art.  169.  A  figure  in  the  first  decimal  place  or  order, 
expresses  a  fraction  whose  denominator  is  a  unit  with  one 
cipher  annexed  ;  thus,  .2  =  j^^  :    and, 

A  figure  in  the  second  decimal  place,  expresses  a  fraction, 
whose  denominator  is  a  unit  with  two  ciphers  annexed ; 

Thus,  .02  =  j^q:    also, 

A  figure  in  the  third  place  expresses  a  fraction  whose 
denominator  is  a  unit  with  three  ciphers  annexed ; 

Thus,   .002  =  -j-QQ-Q  ;  and  so  on  :    hecice. 

The  denominator  of  a  Decimal  Fraction  is  1,  icith  as  many 
ciphers  annexed  as  there  are  decimal  places  in  the  numerator. 

Art.  170.  table  of  decimal  places  or  orders. 


5         •  "   a   2        -go 

~       ii'23.aJ3:3S 
=   2  -^   -S  ^    9 


.  6  is  read  6  tenths. 

.25  25  hundredths. 

.013  13  thousandths. 

.0305  305  ten-thousandth.q. 

.00108  108  hundred-thousandths. 

.378659  378659  millionths. 

5.0000124  5  and  124  ten-millionths. 

2  6.00000005  26  and  5  hundred-millionths. 

Rem. — The  orders  called  ten-thousandths,  hundred-thousandths,  &c. 
are,  really,  tenths  of  thousandths,  hundredths  of  thousandths,  &o. 


DECIMAL    FRACTIONS.  179 

Art.  171.  The  table  shows,  that  the  value  expressed 
by  any  figure  in  decimals^  as  well  as  in  whole  numbers, 
depends  on  the  place  from  units  which  it  occupies  : 

Thus,   .2  expresses  2  tenths,  Ya. 

.02        ..         2  hundredths,        j|q. 

.002     ..         2  thousandths,     j(f(jo' 
But,  Jq  of  f^  is  -^§0  ;     J^  of  y3_  is  -2__^  and  so  on. 

Hence,  as  the  jigures  decrease  from  left  to  right  tenfold,  such 
fractions  are  called  decimals,  from  the  Latin  decein,  meaning  ten. 

ANNEXING    AND    PREFIXING    CIPHERS. 

Art.  172.  Since  the  orders  decrease,  from  left  to  right, 
in  a  tenfold  ratio,  and  each  cipher  prefixed  to  a  decimal 
removes  it  one  place  toward  the  right,  therefore, 

Each  cipher  prefixed  to  a  Decimal,  diminishes  its  value 
ten  times. 

Thus,  prefix  one  cipher  to  the  decimal  .5,  it  becomes  .05 ; 
prefix  two,  it  becomes  .005 ;  three,  .0005 :  each  fraction 
expressing  one-tenth  of  the  preceding. 

As  the  value  expressed  by  a  Decimal  depends  on  the 
place  from  units  which  it  occupies,  and  as  its  place  is  not 
changed  by  annexing  ciphers,  therefore, 

Annexing  ciphers  to  a  Decimal  does  not  alter  its  value. 

Thus,    .  2  =  .  20  = .  200  ;  that  is,  f^  =  ^^o_  ^  _2_o.o_. 

Rem. — 1.  Annexing  ciphers  to  a  decimal,  is  equivalent,  to  multi- 
plying both  terms  of  a  fraction  by  the  same  number,  which  does 
not  alter  its  value.     Art.  135. 

2.  The  effect  of  annexing  and  prefixing  ciphers  to  decimals,  is 
the  reverse  of  that  in  whole  numbers. 

Review. — 1 68.  How  do  the  orders  of  whole  numbers  decrease  from  left 
to  right  ?  Give  an  example.  How  may  the  orders  be  continued  from  the 
place  of  units  toward  the  right  ? 

168.  What  will  the  order  next  to  units  express  ?  The  next  order  ?  The 
next  ?  How  may  y^  be  written  without  the  denominator  ?  j^  ?  yo  ^ 
ihlS  ^  T§?J  ^  7§0  "^  ^^  writing  one-hundredth,  why  place  a  cipher  in 
the  order  of  tenths  ? 

169.  What  is  the  denominator  of  the  fraction,  expressed  by  a  figure  in 
the  first  decimal  place?  In  the  second?  In  the  third?  What  is  the 
denominator  of  a  decimal  ? 


180  RAY'S    PRACTICAL    ARITHMETIC. 

Art.  173.  As  ten  hundredths  make  one  tenth,  and 
ten  tenths  one  unit,  the  decimals  increasing  from  right  to 
left  as  in  whole  numbers,  Art.  8 ;  therefore, 

Whole  numbers  and  decimals  may  he  written  in  the  same 
line,  by  placing  the  separating  point  between  them ;    thus, 

2  units  and  2  hundredths  are  written  2.02 

A  whole  number  and  a  decimal,  written  together,  is  a 
mixed  decimal  number. 

The  decimal  point  ( . ) — called  a  separatrix — separates  whole 
numbers  and  decimals. 

Art.  174.  If,  in  United  States  Money,  the  dollar  be 
taken  as  the  unit,  dimes  will  be  tenths,  cents  hundredths, 
and  mills  thousandths  of  a  dollar  ;    hence. 

Any  sum  in  U.  S.  Money  may  be  expressed  in  dollars 
and  decimal  fractions  of  a  dollar.     Thus, 

$2  and  15cts.,  are  $2,  and  15  hundredths ;    written,  $2.15 

$1  and  8  mills,  are  $1,  and  8  thousandths ;  written,  $1,008 

DECIMAL    NUMERATION    AND    NOTATION. 

Art.  175.  To  Read  Decimals;  First,  read  the  decimal  a^ 
a  whole  number,  then  add  the  name  of  the  right  hand  order. 

Thus,      .6     is  read  six  tenths. 

.06       ..        six  hundredths. 

.204     ..        two  hundred  and  four /^oz/sanrf/ As. 

Rem. — 1.  In  reading  abstract  mixed  decimal  numbers,  for  ex- 
ample, 400.035,  the  word  units  should  be  added  after  the  whole 
number,  to  distinguish  it  from  the  decimal  .435     Thus,  400.035  is 

Review. — 170.  Begin  with  tenths,  and  name  the  places  of  decimals  in 
order.     Rem.  What  is  the  meaning  of  ten-thousandths  ? 

171.  On  what  does  the  value  expressed  by  any  figure  depend  ?  How  do 
decimals  decrease  from  left  to  right  ? 

172.  How  does  each  0  prefixed  affect  the  valuo?  "Why?  Give  examples. 
How  docs  annexing  ciphers  affect  the  value  ?     Why  ?     Give  examples. 

172.  Rem.  1.  Annexing  ciphers  t^)    a  decimal  is  equivalent  to  what? 
2.  How  does  annexing  and  prefixing  ciphers,  compare  with  the  same 
operation  in  whole  numbers  ? 


DECIMAL    FRACTIONS. 


181 


read,  four  hundred  units  and  thirty-five  thousandths ;  while  .485  is 
read,  four  hundred  and  thirty-five  thousandths.  When  the  mixed 
number  is  concrete,  [yards,)  instead  of  saying  400  units,  &c., 
say,  400  yards  and  35  thousandths. 

2.  Another  method :  Read  the  whole  number,  add  the  word  deci- 
mal, then  read  the  fraction.  Thus,  read,  25.025,  twenty-five,  decimal, 
twenty-five  thousandths ;  or,  twenty-five,  decimal,  nought-two-five. 

3.  Before  reading  the  decimal,  let  the  pupil  ascertain  the  name 
of  the  right  hand  order.  Begin  at  the  separatrix,  point  to  each 
order  successively,  and  pronounce  its  name ;  thus,  tenths,  hun- 
dredths, thousandths,  ten-thousandths,  &c. 

EXAMPLES   TO   BE   COPIED   AND   READ. 


(1.)     (2.) 

(3.) 

(4.) 

(5.) 

5     .0003 

.00004 

.000007 

.0000008 

06    .0625 

.00137 

.000133 

.00000009 

003   .2374 

.02376 

.001768 

.1010101 

028   .2006 

.31456 

.040035 

.00100304 

341   .0104 

.01007 

.360004 

.00040005 

(6.) 

(7.) 

(8.) 

6.5 

.6000000 

40504.01037 

60.04 

.0080000 

54000000.000054 

184.173 

.1020000 

30701000.1037025 

Art.  176.  To  Write  Decimals  ;  First,  write  the  number 
of  parts  expressed  by  the  decimal,  as  a  whole  number ;  then  prefix 
ciphers  till  the  right  hand  figure  stands  in  the  required  order. 

Thus,  in  writing  twenty-five  ten-thousandths,  first  write  25, 
then  prefix  two  ciphers,  so  that,  beginning  at  tenths  to  name  the 
orders,  5  may  stand  in  the  order  of  ten-thousandths ;  .0025 


Review. — 173.  How  are  whole  numbers  and  decimals  written  in  the 
same  line  ?     "What  are  they  called  ?     What  the  decimal  point  called  ? 

174.  If  the  dollar  be  taken  as  the  unit,  what  are  dimes  ?  Cents  ?  Mills  ? 
How  express  any  sum  in  U.  S.  Money  ?    Give  examples. 

175.  "What  the  rule  for  reading  decimals  ?  Rem.  1.  How  read  abstract 
mixed  numbers  ?  Why  ?  2.  What  other  method  ?  176.  What  the  rule 
for  writing  decimals  ? 


182 


RAY'S    PRACTICAL    ARITHMETIC. 


Note. — To  change  a  Com.  Fraction  wliose  denominator  is  1,  with 
ciphers  annexed,  as,  j^f^fyy,  to  a  decimal  form,  prefix  so  many 
ciphers  to  the  numerator,  that  the  number  of  places  in  the  decimal 
shall  equal  the  ciphers  in  the  denominator,  Art.  169. 

Express  these  Fractions  and  Mixed  numbers  in  Decimals. 


1. 
2. 


1  0' 


13  "^ 


_5 

1  b  (J' 

24-^77 

■^^1  0(J> 


_1  35 
1  0  0  OU' 

QO    2  5 


2  0  3 
TUbooTJ' 

ft        4 


_ ^4  00.6.^^ 
1  OooooOTJ* 

Q        105 
°lU"0  0'OUO* 


EXEKCISES  IN  WRITING  DECIMALS. 


3.  Four  tenths. 

4.  Two  tenths  and  cix  hun- 
dreaths. 

5.  Thirty-five  hundredths. 

6.  Eight  hundredths. 

7.  Five  thousandths. 

8.  Three  hundred  and  four 

9.  Four  thousand  one  hun- 
dred and  twenty-five  ten-thou- 
sandths. 

10.  Two  hundred  and  five  ten- 
thousandths. 

11.  Eight  ten-thousandths. 

12.  20  thousand  three  hundred 
and  four  hundred-thousandths. 

13.  Six  hundred  and  five  hun- 
dred-thousandths. 

14.  'Sinehundred-thousandths. 

15.  Three    hundred   thousand 
and  four  millionths. 

16.  203  millionths. 

17.  Seven  millionths. 

18.  Twenty-four  ten-millionths. 

Art.    177.    Mixed   decimal   numbers   may  be   read    as 
though  they  were  entirely  decimals. 

Thus,  6.5  =  6  5q  =  f  g  :     it  may  be  read,  G5  tenths. 
Ilence,  when  a  mixed  decimal  number  is  expressed  in  decimals, 


19.  Eighty-thousand   and   six 
ten-millionths. 

20.  Two  hundred-millionths. 

21.  Nine  hundred  and  seven 
hundred-millionths. 

22.  20  million  20  thousand  and 
three  hundred-millionths. 

23.  One  million  ten  thousand 
and  one  hundred-millionths. 

24.  Twenty  units  and  twenty- 
five  hundredths. 

25.  One  hundred  and  six  units 
and  thirty-seven  thousandths. 

26.  One  thousand    units  and 
one  thousandth. 

27.  Two  hundred    units    and 
twenty-five  thousandths. 

28.  29  units  and  29  millionths 

29.  One  million  and  five  hil- 
lionths. 

30.  Two   hundred    units   and 
two  ten-billionths. 

31.  Sixty-five    units    and    six 
thousand  and  five  millionths. 


DECIMAL  FRACTIONS.  183 

it  may  he  written  as  a  whole  number,  and  the  point  so  placed^ 
thai  the  right  hand  Jigure  shall  stand  in  the  required  order. 

32.  One  hundred  and  forty-three  tenths. 

33.  Three  thousand  two  hundred  and  four  hundredths. 

34.  Ten  thousand  five  hundred  and  two  ten-thousandths, 

ADDITION    OF    DECIMALS. 

Art.  178.  1.  Add  4  and  4  ten-thousandths;  28  and 
35  thousandths  ;  8  and  7  hundredths ;  and  9404  hundred- 
thousandths. 

Solution. — Since  only  numbers  of   the  operation. 

same  unit  value  can  be  added,  write  figures  4.0004 

of  the  same  order  under  each  other.  28.035 

Since  ten  units  of  any  order  (Art.  173)  8.07 

make  one  unit  of  the  order  next  higher,  .09404 

add  the  figures  in  each  order,  and  carry  .     ~~A(\~\~CkCiA~I. 
one  tor  every  ten,  as  in  Addition  oi  Simple 
Numbers,  placing  the  separatrix  under  the  points  above. 

*2.  Find  the  sum  of  3  units  and  25  hundredths;  6 
units  and  4  tenths  ;  and  35  hundredths.  Ans.  10. 

Rule  for  Addition. —  Write  numbers  of  the  same  order  under 
each  other ^  tenths  under  tenths^  hundredths  under  hundredths,  &c. 

Then  add,  as  in  Addition  of  Simple  Numbers,  and  point  off  in 
the  amount  as  many  places  for  decimals  as  are  equal  to  the  greatest 
number  of  decimal  places  in  either  of  the  given  numbers. 

Proof. — The  same  as  in  Addition  of  Simple  Numbers. 

3.  Add  21.611  ;  6888.32;  3.4167    Ans.  6913.3477 

4.  Add  6.61;  636.1;  6516.14;  67.1234;  and  5.1233 

Ans.  7231.0967 

Review, — 177.  How  may  mixed  decimals  be  read  ?  Give  an  example. 
When  a  mixed  decimal  is  expressed  in  words,  how  written  ? 

178.  In  writing  decimals  to  be  added,  why  place  figures  of  the  same 
order  under  each  other  ?  Why  carry  one  for  every  ten  ?  Where  place  the 
separating  point  ?    How  prove  Addition  of  decimals  ? 


184  RAYS    PRACTICAL    ARITHMETIC. 

5.  Add  4  and  8  tenths ;  43  and  31  hundredths;  74  and 
19  thousandths;  11  and  204  thousandths.    Ans.  133.333 

6.  Add  45  and  19  thousandths;  7  and  71  hundred- 
thousandths  ;  93  and  4327  ten-thousandths ;  6  and 
401  ten-thousandths.  Ans.  151.49251 

7.  Add  432  and  432  thousandths;  61  and  793  ten- 
thousandths;  100  and  7794  hundred-thousandths;  6.009; 
1000  and  1001  ten-thousandths.  Ans.  1599.69834 

8.  Add  16  and  41  thousandths;  9  and  94  millionths ; 
33  and  27  hundredths;  8  and  969  thousandths;  32  and 
719906  millionths.  Ans.  100. 

9.  Add  204  and  9  ten-thousandths  ;  103  and  9  hun- 
dred-millionths  ;  42  and  9099  millionths  ;  430  and  99  hun- 
dredths; 220.0000009  Ans,  999.99999999 

10.  Add  35  ten-thousandths;  .00035;  35  millionths  and 
35  ten-millionths.  Ans.  .0038885 

fi^*  For  adJitioual  problems,  see  Ray's  Test  Examples. 

STJBTEACTION  OF  DECIMALS. 

Art.  179.  1.  From  20  and  14  thousandths,  subtract 
7  and  21  ten-thousandths. 

Solution.  —  Write  numbers  of  the    same  operation. 

order  under  each  other,  because  the  difference  20.014 

only  between  numbers  of  the  same  unit  value  7.0021 

can  be  found.     Since  annexing  ciphers  to  a  

decimal  does  not  alter   its   value,  Art.  172,      Ans.  lo.Oll  J 
regard  the  place  of  ten-thousandths  in  the  minuend,  as  occupied 
by  a  cipher.     Then  Subtract  as  in  Simple  Numbers. 

*2.  From  5  and  .03  take  2  and  .115  Ans.  2.915 

Rule  for  Subtraction. —  Write   the  less  number  under  the 

greater  J  tenths  under  tenths.^  hundredths  under  hundredths,  ci'c. 

Subtract,   a*  in  Simple  Numbers,  and  point  off  the  decimal ^ 

places  as  in  Addition  of  Decimals. 

Proof. — The  same  as  in  Subtraction  of  Simple  Numbers. 

Review. — 179.    In  Subtraction  of  decimals,  why  aro  tenths  written 
Unilor  tenths,  hundredths  under  hundredths,  <tc.? 


DECLMAL  FRACTIONS.  185 

3.  From  24.0042  take  13.7013  Ans.  10.3029 

4.  170.0035  take  68.00181  Ans,  102.00169 

5.  .0142  take  .005  Ajis.   .0092 

6.  .05  take  .0024  Ans.   .0476 

7.  13.5  take  8.037  Ans.  5.463 

8.  3  take  .00003  Ans.  2.99997 

9.  29.0029  take  19.003  Ans.  9.9999 

10.  5  take  125  thousandths.  Ans.  4.875 

11.  10000  take  1  ten-thousandth.        Ans.  9999.9999 

12.  1  take  1  millionth.  Ans.   .999999 

13.  25  thousandths  take  25  millionths.  Ans.   .024975 

MTJLTIPLICATION  OF  DECIMALS. 

Art.  180.  Multiplication  of  decimals  embraces  two  cases. 

1.  To  multiply  together  a  decimal  and  a  wliole  number. 

2.  To  multiply  together  two  decimals. 

Art.  181.     1.  Multiply  125  thousandths  by  9. 

Solution. —  Regard  125  as  the  numerator  of  a  operation". 

fraction,    the    denominator    being     1000.      Then,  .125 

since  a  fraction  is  multiplied  by  multiplying  its  9 

numerator,  125  X  9  =  1 125  thousandths  =  1 .  125,  by  

proper  pointing  (Art.  177).  ^^^^'  1.125 

In  this  case,  the  number  of  decimal  places  in  the  product 
must  equal  the  number  in  the  multiplicand.  If  9  be  multiplied 
by  .125,  the  product  will  also  be  1.125  (Art  30). 

*2.    Multiply  35  hundredths  by  7.  Ans.  2.45 

*3.    Multiply  2  tenths  by  8  tenths. 

OPERATION.      . 2  X  . 8  =  j2^  X  j^Q = -j\%  =.16  Aus. 
4.    Multiply  2  hundredths  by  4  tenths. 
OPERATION.      .02X.4=-i-§oXy%  =  -j-^8_^=.008  Ans. 
Here,  the  decimals  are   converted   into   common    fractions,  for 

Review. — 179.  How  is  the  subtraction  performed ?  Why?  Give  the 
Rule.  180.  What  two  cases  are  embraced  in  Multiplication  of  decimals  ? 
181.  How  multiply  a  decimal  fraction  by  a  whole  number? 


186 


RAYS    PRACTICAL    ARITHMETIC. 


the  purpose  of  explaining  the  principle  on  which  the  product  is 
pointed. 

The  denominator  of  the  product  of  two  decimals  will  be  1,  with 
as  many  ciphers  annexed  as  there  are  ciphers  in  both  the  de- 
nominators. But  the  number  of  ciphers  in  each  denominator  is 
the  same,  Art.  169,  as  the  number  of  places  in  the  decimal. 

Hence,  the  number  of  decimal  places  in  the  product,  must  equal 
the  number  of  decimal  j)laces  in  both  /actors. 

^5.  Multiply  15  hundredths  by  7  tenths.    Ans.  .105 

General  Rule  for  Multiplication. — Multiply  as  in  Simple 
Numbers ;  point  off  from  the  right  of  the  product  as  many 
figures  for  decimals  as  there  are  decimal  places  in  both  mul- 
tiplicand and  multiplier ;  if  there  be  not  so  many  places  in  the 
product^  supply  the  deficiency  by  prefixing  ciphers. 

Proof. — The  same  as  in  Multiplication  of  Simple  Numbers. 


6. 

Multiply  125.015 

7. 

Multiply             .135 

by                 .001 

by 

.005 

Ans.    .125015 

Ans.   .000675 

8. 

Multiply  1.035  by  1 

7. 

Ans.  17.595 

9. 

19  by  .125 

Ans.    2.375 

10. 

4.5  by  4. 

Ans.     18. 

11. 

.G25  by  64. 

Ans.     40. 

12. 

61.76  by  .0071 

Ajis.  .438496 

13. 

1.325  by  .0716 

Ans.     .09487 

14. 

79000  by  .079 

Ans.     6241 . 

15. 

1  tenth  by  1  hundr 

edth. 

Ans.       .001 

16. 

1  by  1  ten-thousa 

tidth. 

Ans.     .0001 

17 

43  thousandths  by 

.0021 

Ans.  .0000903 

18. 

40000   by  1  millic 

)nth. 

Ans.   .04 

19. 

.09375  by  1  &  G4 

millionths. 

Ans.   .093756 

Art.  182.  The  opcratioi 

IS  of  multi 

plying  by  10,   100, 

IvEViKW. — 181.  To  what  is  the  denominator  of  the  product   of   two 
decimals  equal?     To  what  the  number  of  ciphers  in  each  denominator? 


DECIMAL   FRACTIONS.  187 

1000,  &c.,  may  be  shortened  by  removing  the  decimal 
point  as  many  places  to  the  right^  as  there  are  ciphers 
in  the  multiplier:  and, 

If  there  be  not  so  many  figures  on  the  right  of  the 
point,  annex  ciphers  to  supply  the  deficiency. 

Thus,  2.07X10  =  20.7 

8@°°  For  additional  problems,  see  Ray's  Test  Examples. 

DIVISION  OF  DECIMALS. 

Art.  183.  Decimals  may  be  divided  when  the  divisor, 
or  dividend,  or  both,  are  decimals. 

Since  the  dividend  is  equal  to  the  product  of  the  divisor  and 
quotient,  it  must  contain  as  many  decimal  places  as  there  are 
decimals  in  both  divisor  and  quotient.     Art.  181:  hence. 

There  must  be  as  many  decimals  in  the  quotient  as  the 
decimal  places  in  the  dividend  exceed  those  in  the  divisor. 

1.  Divide  2.125  by  5  tenths. 

Solution.— Divide  as   in   Simple   Numbers;  rw^^^oS' 

then,  since   there  are   three  decimal  places  in  *    )     

the  dividend,  and  one  in  the  divisor,  point  oflF  Ans.  4.25 
two  decimals  in  the  quotient. 

2.  Divide  21  units  by  .5  operation. 
Solution. — In  dividing  a  whole  number  by         .OJ-j1.U 

a  fraction,   the    whole   number  is  reduced   to         ^«s    4*^ 

the  same  parts  of  a  unit  as  the  fraction,  that 

the  divisor  and  dividend  may  be  of  the  same  denomination. 

So,  in  dividing  a  whole  number  by  a  decimal;  reduce  the 
dividend  to  the  same  denomination  as  the  divisor,  by  annexing 
to  it  as  many  ciphers  as  there  are  decimal  places  in  the  divisor, 
and  the  quotient  is  a  whole  number. 

3.  Divide  83.1  by  4. 

Solution. — Divide  the  figures  of  the  dividend  by  the  divisor, 
as  in  whole  numbers,  and  a  remainder  occurs.     Then,  to  continue 

Review. — 181.  To  what  is  the  number  of  decimal  places  in  the  product 
equal  ?    "What  is  the  General  Rule  for  multiplication  ? 

182.  How  multiply  a  decimal  by  10,100,  &c.  ?  183.  When  may  decimals 
be  divided  ?    How  many  decimal  places  must  the  quotient  contain  ?    Why  ? 


188  RAY'S   PRACTICAL   ARITHMETIC. 

the  division,  annex  ciphers  to  the  dividend,  operation. 

which  does  not  alter  its  value,  (Art.  172),  4^83   100 

and  divide  as  before.     Continue  the  divi-  

sion  until  there  is  no  remainder,  or  until  Ans.   20.775 
the  quotient  is  sufficiently  exact. 

As  there  are  three  places  of  decimals  in  the  dividend,  and  none 
in  the  divisor,  there  must  be  three  in  the  quotient. 

4.  Divide  2.11  by  .3 

OPERATION. 

In  this  example,  the  division  will  not  *^^9  11000 

terminate.     In  such  cases,  it  is  to  be  car-  *    -^  "*' 

ried  to  sufficient  exactness:  the  sign -{-is  Aiis.  7.0333-{- 

annexed  to  denote  that  the  division  is  not  complete. 

*5.  Divide  1.125  by  .03  Ans.  37.5 

*6.  Divide  2  by  .008  Ans.  250. 

*  7.  Divide  37 . 2  by  5 .  Ans.  7 .  44 

General  Rule  for  Division. — Divide  as  in  Simple  Nnm- 
bers,  and  point  off  from  the  right  hand  of  the  quotient  as 
many  places  for  decimals  as  the  decimal  places  in  the  dividend 
exceed  those  in  the  divisor;  if  there  he  not  so  many  places, 
supply  the  deficiency  by  prefixiny  ciphers. 

Proof. — The  same  as  in  Division  of  Simple  Numbers. 

Notes. — 1.  When  the  divisor  has  more  decimals  than  the  divi- 
dend, annex  ciphers  to  the  dividend  until  its  decimal  places  equal 
those  of  the  divisor ;  the  quotient  will  be  a  whole  number. 

2.  After  dividing  all  the  figures  of  the  dividend,  if  there  be  a 
remainder,  annex  ciphers  to  it,  and  continue  the  division  till  there 
is  no  remainder,  or  until  the  quotient  is  sufficiently  exact.  In  point- 
ing the  quotient,  regard  the  ciphers  annexed  as  decimal  places. 

8.  Divide  86 .  075  by  27 . 5  Ans.      3.13 

9.  24.73704  by  3.44  Ans.    7.191 
10.       206.166492  by  4.123                    ^«s.  50.004 

Review. — 183.  What  is  tho  General  Rule  for  Division?  Note  1. 
When  tho  number  of  places  in  tho  divisor  exceeds  those  in  tho  dividend, 
what  is  required  ?     Why  ?     What  will  bo  the  quotient  ? 


DECIMAL  FRACTIONS.  1Q9 

11.  Divide  100.8788  by  454.  Ans.     .2222 

12.  .000343  by  3.43  Ans.     .0001 

13.  9811.0047  by  .108649  Ans.  90300. 

14.  .21318  by  .19  Ans.    1.122 

15.  102048  by  .3189      Ans.  320000. 

16.  .102048  by  3189.      ^?is.  .000032 

17.  9.9  by  .0225         Aiis.  440. 

18.  How  often  is  10  contained  in  .1?  Ans.     .01 

19.  1  tenth  contained  in  1?  Ans.    10. 

20.  1  hundredth  in  10?  Ans.  1000. 

21.  64  in  1  and  7  tenths?  Ans.  .0265625 

22.  80  in  8  hundredths  ?  Ans.    .001 

23.  1000  in  1  thousandth?  Ans.    .000001 

24.  Divide  1  thousandth  by  1  thousandth.     Ans.  1 . 

25.  1  ten-thousandth -^  1  ten-millionth,    ^ws.  1000. 

26.  1  hundredth  H-4  millionths.  Ans.  2500. 

27.  1.5^.7  ^ws.  2.142857-f- 

Art.  184.  To  divide  a  Decimal  by  10,  100,  1000,  &c., 
remove  the  decimal  point  as  many  places  to  the  left  as 
there  are  ciphers  in  the  divisor  : 

And,  if  there  are  not  so  many  figures  on  the  left  of  the 
point,  supply  the  deficiency  by  pre/iicw*^  ciphers. 

Thus,  18.3  divided  by  10     =1.83 
18.3  divided  by  100  =  .183 
18.3  divided  by  1000=   .0183 

Art.  185.   EEDTJCTION    OF    DECIMALS. 
Case  I. —  To  reduce  a  common  fraction  to  a  decimal. 

1.  Reduce  the  fraction  |  to  a  decimal. 

Sol. — The  numerator,  3,  will  not  be  changed  by  writing  ciphers 
in  the  place  of  tenths,  hundredths,  &c.  Since  a  fraction  is  ex- 
pressed in  the  form  of  an  unexecuted  division  (Art.  125),  regard 
the  operation  as  division  of  decimals,  and  perform  it  according  to 


190  KAY'S    PRACTICAL   ARITHMETIC. 

the   rule,   (Art.   183).     Since  the  divisor  has  no       operation. 

decimal  places,  the  quotient  must  have  as  many  4)3.00 

places  as  there  are  ciphers  annexed.  .  IT 

^  Ans.    .70 

*2.  Reduce  i  to  a  decimal.  »  ^ns.  .125 

Kule  for  Case  I. — Annex  ciphers  to  the  numerator^  divide 
by  the  denominator,  and  point  off  in  the  quotient  as  many  places 
for  decimals  as  there  are  ciphers  annexed  to  the  numerator. 

Note. — When  common  fractions  can  not  be  exactly  expressed  in 
decimals,  continue  to  divide  till  the  quotient  is  sufficiently  exact. 

REDUCE   THESE   COMMON    FRACTIONS   TO   DECIMALS  : 


3. 

4 
5- 

Ans.   .8 

8. 

■4  00- 

Ans.   .0225 

4. 
5. 

4 

Ans.  .  16 
Ans.  .075 

9. 
10. 

1 

5 

Ans.   .00390625 
Ans.   .8333-1- 

6. 

I  5 

Ans.  .9375 

11. 

4 
3  3' 

Ans.   .121212-h 

7. 

T2d(J- 

Ans.  .0008 

12. 

1 
1  1* 

Ans.   .0909094- 

CAS 

>E  II 

Art.  186.    To  reduce  a  decimal  to  a  common  fraction, 
and  to  its  lowest  terms. 

1.  Reduce  .75  to  a  common  fraction  in  its  lowest  terms. 


Sol. — By  writing  the  denominator,    .75 


OPERATION. 


becomes -'''~\,  and,  reduced   to  its   lowest         ^7  c        -=        ,     j 
terms  by  the  rule  (Art.  138),  it  becomes  |.  i  oo      j 

*2.  Reduce  .6  to  a  common  fraction,  in  its  lowest 
terms.  Ans.  ?. 

Rule  for  Case  II. —  Write  the  denominator  under  the 
decimal ;  it  will  then  be  a  common  fraction,  ichich  reduce  to 
its  lowest  terms,  (Art.  138). 

REDUCE  TO  COMMON  FRACTIONS,  IN  THEIR  LOWEST  TERMS, 

7.  .033  Ans.    yggg. 


3.  .25  Ans.  \. 

4.  .375  Ans.  |. 

5.  4.02  Ans.  4^»5. 

6.  8.415  Ans.Q^}f^. 


8.  .5625  Ans.-^^, 

9.  .34375  Ans.  Ji. 
10.      .1484375      Ans.  -y/^. 


DECIMAL  FRACTIONS.  191 

Note. — When  a  decimal  contains  several  places  of  figures,  their 
Value  can  be  found  nearly  by  inspection.  To  do  this,  take  ih^  first 
or  {\\Q  first  two  figures,  as  the  numerator,  of  which  10  or  100  is  the 
denominator,  then  reduce  it  to  its  lowest  terms: 

Thus  .332125  is  nearly  1  third;  .258321  is  nearly  1  fourth. 

CASE    III. 

Art.  187.  To  reduce  a  decimal  of  one  denomination  to 
an  equivalent  decimal  of  another  denomination. 

1.  Reduce  .25qt.  to  the  fraction  of  a  pt.      operation. 

qt. 
Solution. — To  reduce  quarts  to  pints,  we  multi-  ^25 

fly  by  2,  the  number  of  pints  in  a  quart.  2 

Therefore,    multiply    the  fraction    of    a    quart 

by  2,  to  reduce  it  to  the  fraction  of  a  pint.    Art.  81. 


.50 
Ans.  .5  pt. 


OPERATION. 


2.  Reduce  .5pt  to  tlie  fraction  of  a  qt. 

Solution. — To  reduce  pints  to  quarts,  we  divide  . 

by  2,  the  number  of  pints  in  a  quart.  2  V  5 

Therefore,  divide  ihQ  fraction  of  a  pint  by  2,  to  .           ~Z 

reduce  it  to  the  fraction  of  a  quart.     Art.  81.  ^'*^-  "^^  ^^' 

*3.  Reduce  .125bu.  to  the  fraction  of  a  pk.  Ans.  .5 
*4.  Reduce  .7pk.  to  the  fraction  of  a  bu.  Ans.  .175 

Rule  for  Case  III. — Multiply  or  divide  as  in  Reduction  of 
Tlliole  Numbers,  Art.  81,  according  to  the  rules  for  the  Multi- 
plication and  Division  of  Decimals.     Arts.  181  and  183. 

5.  Reduce  .06251b.  Troy  to  the  frac.  ofanoz.  Ans.  .75 
G.      .05  of  a  yd.  to  the  fraction  of  a  na.  Ans.  .8 

7.     .  00546875  A.  to  the  fraction  of  a  P.      Ans.  .  875 

Review. — 185.  "What  is  Case  1  ?  Eule  for  Case  1  ?  Note.  When  com. 
fractions  can  not  be  exactly  expressed  in  decimals,  what  is  to  be  done  ? 

186.  What  is  Case  2  ?  Rule  for  Case  2  ?  Note.  How  find  the  value 
nearly  of  a  decimal  ?  187.  How  reduce  a  decimal  from  a  higher  to  a  lower 
denomination  ?     From  a  lower  to  a  higher  ? 


192  RAY'S   PRACTICAL  ARITHMETIC. 

8.  Reduce  .0004375  mi.  to  the  frac.  of  a  rd.  Ans.  .14 

9.  .25pt.  to  the  fraction  of  a  gal.  Ans.  .03125 

10.  .6pt.  to  the  fraction  of  a  bu.  Ans.  .009375 

11.  .Smin.  to  the  fraction  of  a  da.    Ans.  .0002083-}- 

12.  .7rd.  to  the  fraction  of  a  mi.         Ans.  .0021875 

CASE    IV. 

Art.  188.  2o  find  the  value  of  a  decimal  in  integers  of 
a  lower  denomination. 

1.  Find  the  value  of  .3125  of  a  bu.  operatio.v. 

bu. 
Solution. — First  multiply  by  4,  as  in   re-  .3123 

ducing  bushels  to  pecks.     Art  81.     This  gives  4 

1  peck  and  .25  of  a  peck ;  then  find  the  value  ,    ~,    o^^Tui 

of  .25  pk.,  by  multiplying  it  by  8,  the  number  ^    '      *    q 

of   quarts   in  a  peck.     This   gives  2  quarts;  

hence,  .3125  of  a  bushel  equals  1  pk.  2qt.  Ans.  qt.  2.00 

*2.  Find  the  value  of  .875  of  a  yd.     Ans.  3qr.  2na. 

Rule  for  Case  IV. — Multiply  the  given  decimal  hy  that 
number  which  will  reduce  it  to  the  next  lower  denomination 
(Art.  81),  and  point  the  product^  as  in  Multiplication  (Art.  181). 

Reduce  this  decimal  in  like  manner ,  and  so  on ;  the  several 
integers  on  the  left  will  be  the  required  answer. 

FIND   THE   VALUE    IN    INTEGERS, 

3.  Of  .7  of  a  lb.  Troy.    .     .     .     Ans.  8oz.  8pwt. 

4.  .8125    ofabu ^72s.  3pk.    2qt. 

5.  .3375    of  an  A ^/?.s.  IR.    14P. 

6.  .04318  of  a  mi.  Ans.  13rd.  4yd.  1ft.  5. 8848 in. 

7.  .33625  of  a  cwt.       .     .     .  Ans.  Iqr.  81b.  10  oz. 

CASE  V. 

Art.  189.  To  reduce  a  quantity  composed  of  one  or  more 
denominations,  to  the  decimal  of  another  quantity  of  one  or 
more  denominations. 


DECIMAL    FRACTIONS.  193 

1.  Reduce  2ft.  Sin.  to  the  decimal  of  a  yard. 

Solution. — By  Art.  163,   the   result   ex-  opkration. 

pressed  as  a  common  fraction,  is  |.  This  2ft.  3  in.  =27  in. 
reduced  to  a  decimal,  becomes  .75,  the  re-  Ijd.  =36  in. 
quired  answer.  f  g  =  |  =  .75 -4»s. 

^2.  Reduce  2pk.  4qt.  to  the  dec.  of  a  bu.    Ans.  .625 

Rule  for  Case  V. — Reduce  the  Jirst  quantity  and  that  of 
which  it  is  to  be  made  a  part^  both  to  the  same  denomination; 
the  less  will  be  the  7iumerator^  and  the  greater  the  denominator 
of  a  common  fraction^  which  reduce  to  a  decimal. 

3.  Red.  13hr.  30min.  to  the  dec.  of  a  da.  Ans.  .5625 

4.  9  dr.  to  the  dec.  of  a  lb.  Av.         Ans.  .03515625 

5.  .028  of  a  P.  to  the  dec.  of  an  A.    Ans.  .000175 

6.  7min.  to  the  dec.  of  a  da.  Ans.  .  0048611 -|- 

7.  4gal.  Iqt.  1 .28pt.  to  the  dec.  of  a  hhd.  Ans.  .07 

8.  What  part  is  3pk.  7qt.  Ipt.  of  2bu.  2pk.  4qt., 
expressed  decimally  ?  Ans.  .  375 

9.  What  part  is  99  pages,  of  a  book  of  512  pages, 
expressed  decimally.  Ans.  .193359375 

10.  A\Tiat  decimal  will  express  the  part  that  55 A.  2R. 
17 P.  is  of  229  A.  2  R.  16 P.?  Ans.  .2421875 

Art.  190.   PROMISCUOUS   EXAMPLES. 

1.  What  cost  9  yards  of  muslin,  at  $0.4  per  yd.,  and  12 
yards  at  §0.1875  per  yd.?  Ans.  §5.85 

2.  What  cost  2.3  vards  of  ribbon,  at  §0.45  per  yd., 
and  1 . 5  yards  at  SO .  375  per  yd.  ?  Ans.  $1 .  5975 

3.  At  82.6875  per  yd.,  what  cost  16  yd.  of  cloth? 
What  16] yd.?  Ans.  to  last,  §43.671875 

4.  At  80.75  per  bushel,  how  much  wheat  can  be  bought 
for  §35.25?  Ans.  47 bu. 

Keview.— 188.  What  is  Case  4  ?   What  the  Rule  fur  Case  4  ?   189.  What 
is  Case  5  ?     What  the  Rule  for  Case  5  ? 
3d  15k  13 


194  RAY'S  PRACTICAL   ARITHMETIC. 

5.  At  $2.5625   per  yard,  how  much  cloth  can   you 
buy  for  $98 . 4  ?  Ans.  38 .4yd. 

6.  AVhat  cost  Gcwt.  2qr.  of  hops,  at  S3. 25  per  cwt.  ? 

SoLU. — Reduce  2qr.  to  the  decimal  of  a  cwt.;  then,  Gcwt.  2qr. 
=G.ocwt.;  §3.25X6.5  =  ^/15.  §21.125 

7.  What  will  be  the  cost  of  7hhd.  23  gal.  of  wine, 
at  849  per  hhd.  ?  Ans.  $360 .  8888-f 

8.  Of  343yd.  3qr.  linen,  at  $.16  per  yd.  ?    ^vis.  e55. 

9.  What  cost  14bu.  3pk.  4qt.  of  corn,  at  $0,625  per 
bushel?  Ans.  $9.296875 

10.  What  will  13 A.  211.  35 P.  of  land  cost,  at  $17.28 
per  acre  ?  A  ns.  $237 .  06 

11.  At  $1.24  per  yard,  how  much  cloth  can  be  bought 
for  $19,065?  Ans.  15.375yd.=15yd.  1  qr.  2na. 

12.  At  $0.3125  per  bu.,  how  much  corn  can  be  bought 
for  $9.296875?  Ans.  29.75bu.=29bu.  3pk. 

13.  At  $4.32  per  A.,  how  much  land  can  you  buy  for 
$59,265?  Ans.  13.71875 A.  =  13A.  2R.  35P. 

14.  Add  .34yd.    .325qr.    .4na.     Ans.  Iqr.  3.14na. 

To  add  or  subtract,  decimals  of  different  denominations,  first  re- 
duce them  to  the  same  denomination.  In  this  example,  reduce  the 
dec.  of  a  yd.  to  qr.,  then  add  the  dec.  of  a  qr. ;  next  reduce  this 
result  to  na.,  and  add  the  dec.  of  a  na. 

15.  From  1 .  53  yards  take  1 .  32  qr.     Ans.  1  yd  3.2  na. 

16.  .05  of  a  year,  (3G5.25  days,)  take  .5  of  an  hour. 

Ans.  18 da.  5hr.  48min. 

17.  .41  of  a  da.  take  .16  of  an  lir. 

Ans.  9hr.  40min.  48 sec. 

18.  In  .4T  .3hhd  .8gal.  howmanypt.?  ^ws.  964pt. 

19.  Find  the  value  of  .3  of  a  year,  (365.25  days,)  iu 
intcG-ers.  Ans.  109  da.  13  hr.  48min. 

20.  In  .005  of  a  year  how  many  sec.  ?     Ans.  157788. 

21.  Whatdecimalof  aC.  islcu.  in?    ^«s.  .000004-1- 


RATIO    OF   NUMBERS.  195 

22  At  $690.35  per  mile,  what  cost  a  road  17 mi.  3 fur. 
15  rd".  long  ?  Ans.  §12027 .  19140625 

In  practice,  only  3  or  4  places  of  decimals  are  generally  used. 
B@^  For  additional  problems,  see  Ray's  Test  Examples. 

XIII.    RATIO. 

Art.  191.  Ratio  is  the  quotient  arising  from  dividing 
one  quantity  by  another  of  the  same  denomination ; 

Thus,  the  ratio  of  2  to  6  is  3 ;    as,  6  -r-  2  gives  the  quotient  3, 
The  ratio  of  2  to  8  is  4  ;    of  2yd.  to  10yds.,  5. 

Illustrations. — 1.  Two  quantities  to  be  compared,  or  to  have 
a  ratio  to  each  other,  must  be  of  the  same  kind,  and  in  the  same 
denomination,  that  the  one  may  be  some  part  of  the  other. 

Thus,  2  yards  has  a  ratio  to  6  yards.  But,  2  yards  has  no  ratio 
to  6  dollars,  the  one  being  no  part  of  the  other. 

2.  Since  ratio  is  the  relation  of  two  numbers  expressed  by  their 
quotient;  and  since  the  quotient  of  2  and  6  may  be  G  divided  by  2, 
or  2  divided  by  6,  either  may  be  used  to  express  their  ratio. 

Thus,  in  comparing  two  lines,  one  of  which  is  2,  the  other  6  inches 
long,  if  the  first  is  taken  as  the  unit  (1)  or  standard  of  comparison, 
the  second  is  three,  that  is,  it  is  3  times  the  first.  If  6  is  taken  as 
the  unit  of  comparison,  2  is  one-third. 

In  finding  the  ratio  between  two  numbers,  the  French  take  the 
first  as  the  divisor,  the  English  the  last.  The  French  method  being 
regarded  the  most  simple,  is  now  generally  used. 

3.  Finding  the  ratio  of  two  numbers,  is  finding  what  part,  or 
what  multiple  one  is  of  the  other. 

The  following  are  equivalent  expressions:  What  is  the  ratio 
of  2  to  6  ?  What  part  of  2  is  6  ?  What  multiple  of  2  is  6  ? 

4.  The  ratio  of  $2  to  $10  is  5 ;  of  $2000  to  $10000  is  also  5 : 
hence,  ratio  shows  only  the  relative  magnitude  of  two  quantities. 

Eeview. — 190,  What  is  ratio  ?  Give  examples,  Illus.  1.  Can 
quantities,  not  of  the  sam-i  kind,  have  a  ratio  ?  Why  not?  2.  What  is 
the  ratio  of  2  to  6  ?     3.  What  arc  equivalent  expressions  ? 

Illus.  4.  What  does  ratio  show  ?  In  finding  the  ratio  between  two  quan- 
tities of  the  same  kind,  but  of  diflferent  denominations,  what  is  required  ? 


196 


RAY'S   PRACTICAL   ARITHMETIC. 


1.  What  is  the  ratio  of  3  to     6? 

2.  Of   2  to  12?        of  3  to  15? 

3.  Of   6  to  18?        of  6  to  30? 

4.  Of    2  to    3  ?       Ans.  #  =  li. 


of  3  to  9'? 
of  7  to  21  ? 
of  5  to  30? 
of  2  to    5  ? 


5.  Of   3  to    4?        of  5  to    8?       of  4  to    6? 

6.  Of  12  inches  to  36  inches  ?        of  2  feet  to  9  feet? 


7.     Of    3  inches  to  1  foot  9  inches  ? 


Ans,  7. 


When    the   quantities   are  of   the  same   kind,   but   of  different 
denominations,  reduce  them  to  the  same  denomination 


8.  What  is  the  ratio  of  3  in.  to  2  ft.  ?     of  4  in.  to  3  yd.  ? 

9.  Of  15  to  25  ?     Ans.  If.       of  25  to  15  ?      Ans.  f . 
10.     Of4tol0?     of  10  to  4?     of6tol6?     of  16  to  6? 


Art.  192.  A  ratio  is  formed  by  two  numbers,  each  of 
which  is  called  a  tei-m,  and  both  together,  a  conpht. 

Thus,  2  and  6  together  form  a  couplet  of  which  2  is  the 
first  term,  and  6  the  second. 

The  first  term  of  a  ratio  is  called  the  antecedent ;  the 
second,  the  consequent. 

Art.  193.  Ratio  is  expressed  in  two  ways 

1st.  In  the  form  of  a  fraction,  of  which  the  antecedent 
is  the  denominator,  and  the  consequent  the  numerator. 

The  ratio  of  2  to  6  is  expressed  by  f ;  of  3  to  12,  by  ^f. 

2d.   By  a  colon  (:)  between  the  terms  of  the  ratio. 

Thus,  the  ratio  of  2  to  6  is  written  2:6;    of  3  to  8,  3 : 8. 

Art.  194.  Since  the  ratio  of  two  numbers  is  expressed 
by  2i,  fraction,  of  which  the  antecedent  is  the  denominator, 
and  the  consequent  the  numerator,  whatever  is  true  of  a 
fraction,  is  true  of  a  ratio.     Hence, 

Revip:w. — 192.  By  what  is  a  ratio  formed?  What  is  each  number 
called  ?     What  both  together  ?    What  is  the  first  term  called  ?    The  2n  ? 

19;^.  In  how  many  ways  is  ratio  expressed  ?  What  is  the  first  method  ? 
The  socoml  ?     Givo  examplos  of  each. 


PROPORTION.  197 

Ist.  To  multiply  the  consequent,  or  divide  the  antecedent, 
multiplies  the  ratio.     Arts.  131  and  133.     Thus, 

The  ratio  of  4  to  12  is  3;     of  4  to  12x5,  is  3x5;  and 
The  ratio  of  4^-2  to  12,  is  6,  which  is  equal  to  3X2. 

2d.  To  divide  the  consequent  or  multiply  the  antecedent,  di 
vides  the  ratio.     Arts.  132  and  134.     Thus, 

The  ratio  of3  to  24  is  8;    of  3  to  24-^2,  is  4,= 8-~2;  and 
The  ratio  of  3  x2  to  24,  is  4,  which  is  equal  to  8-^2. 

3d.  To  multiply  or  divide  both  consequent  and  antecedent  by 
the  same  number,  does  not  alter  the  ratio.    Arts.  134, 135,    Thus, 

The  ratio  of  6  to  18,  is  3;     of  6X2  to  18X2,  is  3;  and 
The  ratio  of  6-^2  to  18-^2,  is  3. 

Art.  195.  A  single  ratio,  as  2  to  6,  is  a  simple  ratio. 

A  compound  ratio  is  the  product  of  two  or  more  simple 

ratios.     Thus, 

The  ratio  Lo  multiplied  by  the  ratio  f,  is  ^fXi  =  ^%  =  ^' 
In  this  case,  3  multiplied  by  5,  is  said  to  have  to  10X6,  the 

ratio  compounded  of  the  ratios  of  3  to  10  and  5  to  6. 

Art.  196.  Ratios  may  be  compared  with  each  other, 
by  reducing  to  a  common  denominator  the  fractions  by 
which  they  are  expressed :  thus, 

To  find  the  greater  of  the  two  ratios,  2  to  5,  and  3  to  8, 
we  have  |  and  |,  which,  reduced  to  a.  common  denominator, 
are  ^g^  and  ^^ ;  and,  as  L^  is  less  than  ^^,  the  ratio  of  2  to  5, 
is  less  than  the  ratio  of  3  to  8. 


XIV.    PROPORTION. 

Art.  197.  Proportion  is  an  equality  of  ratios.  Four 
numbers  are  proportional,  when  the  first  has  the  same 
ratio  to  the  second  that  the  third  has  to  the  fourth. 

Review. — 194.  How  is  a  ratio  aflFected  by  multiplying  the  couse- 
quent,  or  dividing  the  antecedent?  By  dividing  the  consequent,  or 
multiplying  the  antecedent?  By  multiplying  or  dividing  both  consequent 
and  antecedent  by  the  same  number  ?     Why  ?     Illustrate  each. 


198  RAY'S    PRACTICAL    ARITHMETIC. 

Thus,  the  two  ratios,  2  :  4  and  3  :  6,  form  a  proportion, 
since  3  =  5,  each  being  equal  to  2 

Art.  198.  proportion  is  written  in  two  ways  : 

1st.  By  placing  a  double  colon  between  the  ratios. 
Thus,        2:4::     3:6. 

2d.  By  placing  the  sign  of  equality  between  them. 
Thus,        2:4    =     3:6. 

The  first  is  read,  2  is  to  4  as  3  is  to  6 ;  or,  2  has  the  same 
ratio  to  4,  that  3  has  to  6.  The  second  is  read,  the  ratio 
of  2  to  4  equals  the  ratio  of  3  to  6. 

Rem. — 1.  The  least  number  of  terms  that  can  form  a  proportion 
is /owr,  since  there  are  two  ratios  each  containing  two  terms. 

But,  one  of  the  terms  in  each  ratio  may  be  the  same ;  thus, 
2 :  4 : :  4 :  8.  The  number  repeated  is  called  a  mean  proportional 
between  the  other  two  terms. 

2.  The  terms  ratio  and  proportion  are  often  confounded  with  each 
other.  Two  quantities  having  the  same  ratio  as  3  to  4,  are  impro- 
perly  said  to  be  in  the  proportion  of  3  to  4.  A  ratio  subsists 
between  two  quantities;  a  proportion  only  between /our. 

Art.  199.  The  first  and  last  terms  of  a  proportion  are 
called  the  extremes;  the  second  and  third,  the  means. 

Thus,  in  the  proportion  2  :  3  :  :  4  :  6,  2  and  6  are  the 
extremes,  and  3  and  4  the  means. 

Art.  200.  In  every  proportion^  the  product  of  the  mean& 
is  equal  to  the  product  of  the  extremes. 

Illustration.s. — If  we  have  3:4: 
the  ratios  of  each  couplet  being  equal ) 
(Art.  206),  we  must  have i 

Reducing  these  fractions  to  a  com- 
mon denominator  (Art.  155),  gives     . 

Review. — 195.  What  is  a  simple  ratio?  A  compound  ratio?  Give 
examples  of  each.  196.  IIow  compare  ratios  with  each  other  ?  197.  What 
is  proportion  ?     When  are  four  numbers  proportional  ?     Give  examples. 

198.  IIow  is  proportion  written?  IIow  is  the  first  read?  The  second? 
Rem.  1.  What  the  least  number  of  terms  that  can  form  a  proportion? 
2.  What  of  the  terns  ratio  and  proportion  ? 


6   :   8, 

4_8 
3     6 

4X6_ 
18 

3X8 
18 

I 


SIMPLE  PROPORTION.  199 

The  denominators  being  the  same,  and  the  values  of 
the  fractions  being  equal,  the  numerators  must  be  equal ; 
that  is,  4X6,  the  product  of  the  means,  must  =  3x8,  the 
product  of  the  extremes. 

Rem. — The  px-eceding  shows  that  four  numbers  are  not  in  pro- 
portion, when  the  product  of  the  extremes  is  unequal  to  the  pro- 
duct of  the  means.  Thus,  2,  3,  5,  and  8,  are  not  in  proportion, 
for  3X5  is  not  equal  to  2x8. 

Proportions  have  numerous  properties,  the  full  discussion  of 
which  belongs  to  Algebra.     See  "Ray's  Algebra,  First  Book." 

Proportion  is  divided  into  Simple  and  Compound. 

Art.  201.   SIMPLE    PROPORTION. 

Simple  Proportion  contains  only  simple  ratios,  Art. 
195.  It  is  sometimes  called  the  Rule  of  Three,  as 
three  terms  are  given  to  find  a  fourth. 

Rem. — Some  authors  divide  Proportion  into  direct  and  inverse  j  a  dis- 
tinction of  no  utility,  and  always  embarrassing  to  the  learner. 

Art.  202.  Since  the  product  of  the  means  equals  the 
product  of  the  extremes,  and  the  product  of  two  factors 
divided  by  either  of  them,  gives  the  other.  Art.  37, 

Therefore,  If  the  product  of  the  means  he  divided  hy  one  of  the 
extremes^  the  quotient  will  he  the  other  extreme.     Or, 

If  the  product  of  the  extremes  he  divided  hy  one  of  the  means, 
the  quotient  will  he  the  other  mean. 

Thus,  in  the  proportion   2   :    3    :  :   4   :  6, 

3  X  4  -T-  2  =  6,  one  of  the  extremes. 
3  X  4  H-  6  =  2,  the  other  extreme. 
2X6-9-3  =  4,  one  of  the  means. 
2  X  6  -T-  4  =  3,  the  other  mean. 

— • —  ■  '■«- 

Review.— 199.  What  are  the  first  and  last  terms  called?  The  second 
and  third  ?  200.  If  four  numbers  arc  proportional,  to  what  is  the  product 
of  the  means  equal  ?  Eem.  When  are  four  numbers  not  in  proportion  ? 
How  is  proportion  divided  ? 

201.  What  does  Simple  Proportion  contain  ?  What  is  it  called  ?  Why  ? 
202.  When  3  terms  of  a  proportion  are  given,  how  find  the  fourth  ? 


200  RAY'S   PKACTICAL    ARITHMETIC. 

Hence,  If  any  three  terms  of  a  proportion  are  given,  the  fourth 
may  be  found  by  multiplying  together  the  terms  of  the  same 
name,  and  dividing  their  product  by  the  other  given  term. 

1.  The  first  three  terms  of  a  proportion  are  2,  8,  and 
6  :  what  is  the  fourth  term  ? 

Solution. — The    preceding    shows  operation. 

that  the  4th   term  will  be  found  by  2   :   8   :  :   6 

taking  the  product  of  the  2d  and  3d        8X6 
terms,  and  dividing  by  the  1st. 

Or,   by    the    nature   of   proportion, 
Art.  197,  the  ratio  of  the  3d  term  to  the 


=  24.  Ans. 


2 

Or, 


4th  =  the  ratio  of  the  1st  to  the  2d.         f  =  4,  and  6  X  4  =  24. 

Hence,  If  the  third  term  be  multiplied  by  the  ratio  of  the 
first  to  the  second,  the  product  will  be  the  fourth  term. 

EXA3IPLES   TO   BE   SOLVED   BY    EITHER   METHOD. 

2.  The  first  three  terms  of  a  proportion  are  5,  7,  and 
10  :  what  is  the  fourth  term?  Ans.  14. 

3.  The  last  three  terms  are  8,  6,  and  16:  what  is  the 
first  ?  Ans.  3. 

4.  The  first,  third,  and  fourth  terms,  are  5,  6,  and  12 : 
what  is  the  second?  Ans.  10. 

5.  The  first,  second,  and  fourth  terms,  are  3,  7,  and  14: 
what  is  the  third?  Ans.  6. 

6.  Seven  is  to  14,  as  9  is  to  what  number  ?     Ans.  18. 

Art.  203.  1.  If  21b.  of  tea  cost  S4,  at  the  same  rate, 
what  will  be  the  cost  of  G  lb.  ? 

Solution. — Two  pounds  have  the  same   ratio  operation. 

to  61b.,  that  the  cost  of  21b.   ($4),  has  to  the  lb.     lb.       $ 

cost  of  G  lb.     Therefore,  the  first  three  terms  are  2   :    6   :  :  4 

given,  to  find  the  fourth  (Art.  202).  6 

To  find  the  result,  multiply  the  3d  term  by  the  9\o ± 

2d,  and  divide   by  the   1st;  or,  multiply  the  3d  ""-^^ 

term  by  the  ratio  (3)  of  the  first  to  the  2d.  Ans.  812. 

In    stating   this   question,    (arranging  the   terms,)   we 

Review. — 202.  If  the  third  term  of  a  proportion  be  multiplied  by  the 
ratio  of  the  first  to  the  second,  what  will  be  the  product  ? 


SIMPLE    PROPORTION.  201 

may,  with  equal  propriety,  place  the  $4  as  the  first  term. 

Thus,      $4   :    2(1  term   : :    2  lb.    :    6  lb. 

The  operation  of  finding  the  2d  term  in  this  arrange- 
ment, is  like  that  of  finding  the  4th  in  the  preceding. 

It  is  more  convenient  to  arrange  the  terms  so  that  the 
required  term  shall  be  the  fourth;  then, 

Since  a  ratio  can  subsist  only  between  quantities  of  the 
same  denomination,  Art.  191,  Illus.  1, 

The  third  term  of  a  proportion  must  he  of  the  same  denomi- 
nation as  that  in  ivhich  the  answer  is  required. 

2.  If  3  men  can  dig  a  cellar  in  10  days,  in  how  many 
days  will  5  men  dig  it? 


Solution. — Since   the   answer  is   to   be  days^  operation. 

write  10  days  for  the  3d  term.     The  result  will  '^-    i^i-       ^^' 

be  found  by  multiplying  the  3d  term  by  the  2d,  O  :  o  :  :  10 

and  dividing  by  the  1st  (Art.  201);  and,  ^ 

Since  it  will  require  5  men  a  less  number  of  Pj^^O 

days  than  3  men,  place  the  less  number  of  men  

for  the  2d  term,  and  the  greater  for  the  1st.  Ans.  6. 

3.  If  3  men  can   dig   a  cellar  in    10  days,  how  many 
men  will  dig  it  in  G  days? 

Solution. — Since   the    answer   is   to   be  men,  statement. 

write   3  men  for  the  third  term.     Since  it  will  da.    da.       m. 

require  o,  greater  number  of  men  in  6  days  than  6  :  10  :  :  3 

in  10  days,  place  the  greater  number  of  days  for  ,        _ 
the  second  term,  and  the  less  for  the  first. 

*4.  If  3  yd.  cloth  cost  $8,  what  cost  6  yd.?  Ans.  S16. 
*5.  If  5bl.  flour  cost  $30,  what  cost  3  bl.  ?  Ans.  $18. 

GENERAL  RULE  FOR  SIMPLE  PROPORTION. 

1.  Write  for  the  third  term  that  number  which  is  of  the 
denomination  required  in  the  answer. 

2.  If  from,  the  nature  of  the  question^  the  answer  should  he 
greater  than  the  third  term,  place  the  greater  of  the  other  two 
numhers  for  the  second  term,  and  the  less  for  the  first;  but, 


202  RAY'S    PRACTICAL    ARITHMETIC. 

If  the  answer  slioiild  he  less  than  the  third  term,  place  the 
less  number  Jor  the  second  term,  and  the  greater  for  the  first. 

3.  Multiply  the  third  term  hy  the  second,  and  divide  the 
product  hy  the  first ;  the  quotient  will  he  the  ansioer  in  the  de- 
nomination of  the  third  term. 

Notes. — 1.  If  the  first  and  second  terms  contain  difiFerent  denom- 
inations, reduce  them  to  the  same ;  if  the  third  term  consists  of 
more  than  one  denomination,  reduce  it  to  the  lowest  given, 

2.  Multiplying  the  2d  and  3d  terms  together,  and  dividing  by 
the  1st,  is  really  multiplying  the  3d  by  the  ratio  of  the  1st  to  the  2d. 

It  is  most  convenient,  especially  when  the  ratio  is  a  whole  num- 
ber, to  express  it  in  its  simplest  form  before  multiplying. 

3.  After  dividing,  if  there  be  a  remainder,  reduce  it  to  the  next 
lower  denomination,  and  divide  again,  and  so  on. 

6.  If  31b.  12oz.  tea  cost  S3. 50,  what  cost  11  lb.  4oz.? 

Solution.— State  the  question;  and,      „  operation. 

1  .1      1   X        1  oi  .  •      .1  lb.   oz.       lb.   oz.       $ 

to  express  the  1st  and  2d  terms  in  the       o   19.11    4--^^0 

same  denomination,  reduce  both  to  oz. 

Then  multiply  the  third  term  ($3.50)  (KCS  ■  1  S!i(\  •  -  "^  ^0 

by  180,  and  divide  the  product  by  60.  p^^  •  A  P  J^  •  •  ^-^^ 

But  it  is  shorter  to  multiply  at  once  '  

by  3,  the  ratio  of  60  to  180.  Ans.  S10.50 

7.  If  21b.  8oz.  of  tea  cost  $2,  what  quantity  can  you 
buy  for  S5  ?  Am.  61b.  4oz. 

Reduce  the  3d  term  to  ounces  ;  tlie  answer  will  then  be  in  ounces. 

8.  If  4  hats  cost  $14,  what  cost  10  hats?     Ans.  835. 

9.  If  3  caps  cost  69  cents,  what  cost  11  ?    Ans.  82.53 

10.  If  4yd.  cloth  cost  S7,  what  cost  9?      Ans.  815.75 

11.  If  8yd.  cloth  cost  832,  what  cost  12  ?      Ans.  848. 

Review. — 203.  In  stating  a  question  in  simple  proportion,  what  num' 
ber  is  put  for  the  3d  term,  Rule  ?  How  are  the  other  two  numbers  arranged  ^ 
How  is  the  answer  obtained  ?     Of  what  denomination  is  the  answer? 

203.  Note  1.  If  the  first  and  second  terms  contain  different  denomina- 
tions, what  is  required  ?  What  if  the  third  term  contains  more  than  one 
denomination  ?     What  other  explanations  in  Notes  2  and  3  ? 


SIMPLE   PROPORTION.  203 

12.  If  12yd.  cloth  costS48,whatcost8yd.?   Ans.$32. 

13.  If  S32  purcliase  8  yards  of  cloth,  how  many  yards 
will  $48  buy?  '  ^ns.  12yd. 

14.  If  848  purchase  12  yd.  of  cloth,  how  many  yards  can 
be  bought  for  $32  ?  A71S.  8  yd. 

15.  A  man   receives  $152  for  19  months'  work  :  how 
much  should  he  have  for  4  mon.  work  ? 

o  o-  ii  1       i       r-     ii.  OPERATION. 

Suggestion. — Since  the  product  of   the 
second  and  third  terms  is  to  be  divided  by  1  Q  •  4-  •  •  1  (^  *? 

the  first,  therefore,  (Art.  71,)  when  the  first  o 

term  contains  one  or  more  factors  common  t  '^  9  X  4       ^ 

to  either  of  the  other  terms,  shorten  the =  32  Ans. 

operation  by  canceling  the  common  factors.  J.  p 

16.  If  8  men  perform  a  piece  of  work  in  24   days,  in 
what  time  can  12  men  perform  it  ?  Ans.  16  days. 

17.  If  60  men  perform  a  piece  of  work  in  8  days,  how 
many  men  will  perform  it  in  2  days?  Ans.  240. 

18.  If  15  oz.  of  pepper  cost  25cts.,  what  cost  6  lb.  ? 

c  T       ^        1         r.  OPERATION. 

Suggestion.  —  Instead    of 

multiplying  the  6  lb.  by  16,  to  -•  1:^    ,    a    .  .    9f{ 

reduce  it  to  ounces,  indicate  the  -ir     .    pvIT    ••    9^ 
multiplication,  and  cancel  the 

factors  common  to  the  divisor  ^  V  1  6  V  2  ;^ 

anddividend.  In  this  manner,      — — ^160  cts.  Ans. 

operations  may  be  shortened.  %^it 

19.  If  6  gal.  of  molasses  cost  65cts.,  what  cost  2hhd.  ? 

Ans.  $13.65 

20.  If  5cwt.   3qr.  101b.   of  sugar  cost  $21.06,  what 
will  35cwt.  Iqr.  cost?  Ans.  $126.90 

21.  If  1yd.  2qr.  of  cloth  cost  $2.50,  what  will  be  the 
cost  of  1  qr.  2  na.  ?  Ans.  $0 .  62  5 

22.  If  90bu.  of  oats  supply  40  horses  6  days,  how  long 
will  450  bu.  supply  them  ?  Ans.  30 da. 

23.  If  6  men  build  a  wall  in  15  days,  how  many  men 
can  build  it  in  5  days  ?  Ans.  18  men. 


204  RAY'S   PRACTICAL    ARITHMETIC. 

24.  If  15  bu.  of  corn  pay  for  60  bu.  of  potatoes,  how 
much  corn  can  be  had  for  140  bu.  potatoes  ?  Ans.  35  bu. 

25.  If  3cwt.  1  qr.  of  sugar  cost  §22.60,  what  will  be 
the  cost  of  16  cwt.  1  qr.  ?  Ans.  8113. 

26.  If  a  perpendicular  staff,  3  ft.  long,  cast  a  shadow 
4  ft.  6  in.,  what  is  the  height  of  a  steeple,  whose  shadow 
measures  180  ft.  ?  Aiis.  120  ft. 

27.  If  a  man  perform  a  journey  in  60  days,  traveling  9 
hours  each  day,  in  how  many  days  can  he  perform  it  by 
traveling  12  hours  a  day?  Ans.  45  da. 

28.  A  merchant  failing,  paid  60  cts.  on  each  dollar  of 
his  debts.  He  owed  A  $2200,  and  B  S1800:  what  did 
each  receive  ?  Am.  A  S1320.     B  S1080. 

29.  A  merchant  havinj?  failed,  owes  A  8800.30;  B 
8250;  C  8375.10;  D  8500;  F  8115.  His  property, 
worth  8612.12,  goes  to  his  creditors  :  how  much  will  this 
pay  on  the  dollar  ?  Ans.  30  cts. 

30.  If  the  4  cent  loaf  weigh  9  oz.  when  flour  is  88  a  bl., 
what  will  it  weigh  when  flour  is  W  a  bl.?      Ans.  12  oz. 

31.  I  borrowed  8250  for  6mon.  :  how  long  should  I 
lend  8300,  to  compensate  the  favor  ?  Ans.  5  mon. 

32.  A  starts  on  a  journey,  and  travels  27  miles  a  day; 
7  days  after,  B  starts,  and  travels  the  same  road,  36  mi.  a 
day:  in  how  many  da.  will  B  overtake  A?     Ans.  21  da. 

33.  If  William's  services  are  worth  815|  a  mon.,  when 
lie  labors  9  hr.  a  da,,  what  ought  he  to  receive  for  4|  mon., 
when  he  labors  12  hr.  a  da.?  Ans.  891.911 

34.  K  5  lb.  of  butter  cost  8§,  what  cost  |  lb.  ?  Ans.  83^. 

35.  If6yd.clothcost85f,whatcost7^yd.?^7Js.  86y. 

36.  If  J  bu.  wheat  cost  8J,  what  cost  ^  bu.?  Ans.  87^5. 

37.  If  If  yd.  cloth  cost  8^^,  what  cost  2 yd.?  Ans.  8J. 

38.  If  829|  buy  59.\  yards  of  cloth,  how  much  will 
831  j  buy?  '  ^«s.  62.Vyd. 

39.  If  .85  of  a  gallon  of  wine  cost  81.36,  what  will  be 
the  cost  of  .25  of  a  gallon?  Ans.  80.40 

40.  If  61.31b.  of  tea  cost  844.9942,  what  will  be  the 
cost  of  1 .  07  lb.  ?  A  ns.  80 .  78538 


SIMPLE    PROPORTION.  205 

41.  If  I  of  a  yard  of  cloth  cost  $|,  what  will  j\  of  an 
Ell  English  cost?  Ans.  ^^l. 

42.  If  I  of  a  yd.  of  velvet  cost  S4?,  what  cost  17|  yd.  ? 

Ans.  $178,381 

43.  A  wheel  has  35  cogs ;  a  smaller  wheel  working  in 
it,  26  cogs  :  in  how  many  revolutions  of  the  larger  wheel, 
will  the  smaller  gain  10  revolutions?  Aiis.  28|. 

44.  If  a  grocer,  instead  of  a  true  gallon,  use  a  measure 
deficient  by  1  gill,  what  will  be  the  true  measure  of  100 
of  these  false  gallons  ?  Ans.  96|  gal. 

45.  If  the  velocity  of  sound  be  1142  feet  per  sec,  and 
the  number  of  pulsations  in  a  person  70  per  min.,  what 
the  distance  of  a  cloud,  if  .20  pulsations  are  counted  be- 
tween the  time  of  seeing  a  flash  of  lightning  and  hearing 
the  thunder?  Ajis.  3  mi.  5  fur.  145  yd.  2\-  ft. 

46.  The  length  of  a  wall,  by  a  measuring  line,  was  643 
ft.  8  in.,  but,  the  line  was  found  to  be  25  ft.  5.25  in.  long, 
instead  of  25  feet,  its  supposed  length :  what  the  true 
length  of  the  wall?  Ans.  654ft.  11.17  in. 

Note. — A  bushel  of  wheat  is  GO  lb.;  of  rye,  561b.;  corn,  561b.; 
oats,  321b.;  barley,  48  lb. ;  timothy  seed,  421b.;  clover  seed,  GO  lb.; 
flax  seed,  56  lb.     See  Dry  Measure,  Note  2,  page  83. 

The  simplest  method  for  the  following  examples,  is  by  Simple 
Proportion.  The  Jirst  term  will  be  the  number  of  pounds  to  the 
bushel ;  second,  the  weight ;   third,  the  price. 

What  will  be  the  cost  of 

47.  2136  lb.  wheat,  at  75  cts.  per  bu.?      Ans.  $26.70 

48.  1225  lb.  wheat,  at  81  cts.  per  bu.?  .4«s.  116.53+ 

49.  10001b.  rye,      at  63  cts.  per  bu.?      .4/*s.$11.25 

50.  30001b.  oats,     at  24 cts.  per  bu.?      ^ns.  $22.50 

Art.  204.  OF    CAUSE    and    effect. 

The  rule  for  Simple  Proportion  may  be  presented 
under  another  form,  based  on  the  obvious  truth,  that. 

When  a  cause  produces  a  certain  Effect,  if  the  Cause  he  in- 
creased or  diminished,  the  Effect  will  he  increased  or  diminished 
in.  the  same  ratio. 


206  RAYS    PRACTICAL  ARITHMETIC. 

Thus,  if*  the  cause  be  doubled,  the  effect  will  be  doubled,  if 
trebled,  the  effect  will  be  trebled.     Hence  this 

Principle. — Any  cause  is  to  another  similar  cause,  as  the 
EFFECT  of  the  Jirst  cause  is  to  the  effect  of  the  second  cause. 

Illustration. — If  3  men  make  6  rods  of  road  in  a 
day,  how  many  rods  can  4  men  make  ? 

3  men.     :     4  men.     :  :     C  rods.     :     8  rods.    Ans, 
1st  cause.      2J  cause.        1st  effect.     2d  effect. 

The  effect  of  the  2d  cause,  being  the  4th  term  of  a 
proportion,  is  found  by  multiplying  the  3d  term  by  the 
2d,  and  dividing  the  product  by  the  1st.     Art.  202. 

Note. — In  reviewing,  the  questions  may  be  stated  by  this 
principle  ;  or  it  may  be  used  at  first  instead  of  the  Rule,  Art.  203. 

Art.  205.    COMFOUND    PROPORTION. 

A  Compound  Proportion  contains  one  or  more  com- 
pound ratios.     Art.  195. 

2   '   6 ) 
Thus,       A    '.    o\  '•'•    7    :   42,  is  a  compound  proportion. 

Or,  2X4   :   6X8   : :   7   :   42. 

This,  sometimes  called  the  Double  Rule  of  Three,  is  applied 
to  the  solution  of  questions  requiring  more  than  one  statement 
in  Simple  Proportion. 

1.  If  2  men  earn  $20  in  5  days,  what  sum  can  G  men 
earn  in  10  days? 

Here,  the  sum  earned  depends  on  two  things:  the  number  of  men, 
and  the  number  of  days;  that  is,  on  the  ratio  of  2  men  to  G  men, 
and  on  the  ratio  of  5  days  to  10  days. 

First  find  the  sura  earned  by  G  men  in  the  same  time  with  2  men. 
men.  men.        $  $ 

2    :    G    :  :    20   :    60^=  sum  earned  by  G  men  in  5  days. 

Knowing  the  sum  earned  in  5  days,  the  sum  earned  in  10  days 
can  be  found  thus: 
days.  days.         $  $ 

5    :    10   :  :    GO    :    120  =  sum  earned  by  G  men  in  10  days. 

Review. — 201.  On  what  is  Cause  and  Effect  based?  What  ia  tho 
Princii»lu  ?     IImw  find  tho  effect  of  the  2d  cause  ? 


COMPOUND  PROPORTION.  207 

By  examining  these  proportions,  it  is  seen  that  the  ratio  of  the 
third  term  to  the  fourth,  depends  on  two  ratios — 

1st,  the  ratio  of  2  men  to  6  men ;  2d,  the  ratio  of  5  da.  to  10  da. 

The  first  ratio  is  3,  and  the  second  2;  their  product,  3X2^=6, 
is  the  ratio  of  the  third  term  to  the  required  term.     Hence, 

The  ratio,  which  the  3d  term  has  to  the  4th,  is  compounded  of  the 
ratios  of  2  to  6  and  of  5  to  10:  write  the  two  simple  proportions 

Thus,        2  men   :     6  men  ]   . .   ,Z>,   .   ^      ,^. 

5  days  :    10  days  \  "    ^^   '   f^^^th  term. 

Or,  2  X  5   :   6  X  10    : :   20   :   fourth  term. 
20X6X10  =  1200;  2X5=^10;  and  1200 h- 10  =1: §120.   Ans. 

*2.  If  a  man  travel  24  mi.  in  2  da.,  by  walking  4  lir. 
a  day;  at  the  same  rate,  how  far  will  he  travel  in  10  da., 
walking  8  hr.  a  day?  Ans.  240  mi. 

Rule  for  Compound  Proportion.— 1.  Write  for  He  M  term, 

that  number  which  is  of  the  denomination  required  in  the  ansiver. 

2.  Take  any  two  numbers  of  the  same  kijid,  and  arrange  them 
as  in  Simple  Proportion.     Art.  203,  Rule  2,  page  201. 

3.  Arrange  any  other  two  numbers  of  the  same  kind,  in  like 
manner,  till  all  are  used. 

4.  Multiply  the  third  term  by  the  continued  product  of  the 
second  terms ;  divide  the  result  by  the  continued  product  of  the 
first  terms  :  the  quotient  will  be  the  fourth  term,  or  answer. 

« 

Notes. — 1.  If  the  terms  in  any  couplet  are  of  different  denomi- 
nations, reduce  them  to  the  same.  If  the  third  term  consists  of 
more  than  one  denomination,  reduce  it  to  the  lowest  named. 

2.  The  examples  may  be  solved  by  two  or  more  statements  in 
Simple  Proportion,  or  by  Analysis,  Art.  268.  Also,  by  Cause  and 
Effect,  Art.  204,  see  statement,  page  208. 


Review. — 205.  WTiat  does  Compound  Proportion  contain?  To  what 
is  it  applied  ?  In  stating  a  question,  what  number  is  put  for  the  third 
term,  Rule  ?     How  are  the  other  numbers  arranged  ? 

205.  How  is  the  fourth  term  found?  Note  1.  If  the  terms  in  any 
couplet  are  of  diiferent  denomination?,  what  is  required?  What  if  the 
third  term  contains  more  than  one  denomination? 


208  RAYS  PRACTICAL   ARITHMETIC. 

3.  If  6  men,  in  10  days,  build  a  wall  20ft.  long,  3ft, 
high,  and  2  ft.  thick  :  in  how  many  days  could  15  men 
build  a  wall  80ft.  long,  4ft.  high,  and  3ft.  thick? 

STATEMENT  BY  CAUSE  AND  EFFECT. 


1st  cause. 

2d  c.iuse. 

1st  effect. 

2d  effect 

6       : 

15       : 

:       20 

80 

10 

X 

3 

4 

2 

3 

OPERATION    BY  CANCELING. 

Men        ^  l^ 

0^ 

Length    .    Jfp 

?P  ^ 

Hight      .       ^ 

i 

Thickness     ^ 

3     2 

Days  .     .      X 

19^ 

The  terms  of  the  2d  cause  and  1st  effect  form  the  divisors  ;  those 
of  the  1st  cause  and  2d  effect,  the  dividends.  The  x  shows  the 
place  of  the  required  term. 

Explanation.  —  When  the  terms 
forming  ihe  divisor  and  the  dividend, 
contain  one  or  more  common  factors, 
shorten  the  operation  by  Cancellation. 

In  such  cases,  arrange  the  divisors 
on  the  left  of  a  vertical  line,  and  the 
multipliers  [dividends),  on  the  right. 

4X4X2  =  32  da.  Ans. 

4.  If  16  men  build  18  rods  of  fence  in  12  days,  how 
many  men  can  build  72  rd.  in  8 da.  ?  Aim.  96  men, 

5.  If  6  men  spend  $150  in  8mon.,  how  much  will  15 
men  spend  in  20mon.  ?  Ans.  $937.50 

6.  I  travel  217  mi.  in  7  days  of  6  hr.  each,  how  ftir  can 
I  travel  in  9  da.  of  11  hr.  each  ?  Ans.  51Hmi. 

7.  If  $100  gain  $6  in  12  months,  what  sum  will  $75  gain 
in9mon.?  ^«s.  $3,375 

8.  If  1001b.  be  carried  20 mi.  for  20cts.,  how  far  may 
10100 lb.  be  carried  for  $60.60  ?  Ans.  60 mi, 

9.  To  carry  12cwt.  3qr.  400mi.,  costs  S57.12  :  what 
will  it  cost  to  carry  lOtuns  75 mi.?  Ans.  $168. 

10.  If  18  men,  in  15  da.,  build  a  wall  40 rd.  long,  5ft. 
lii'jjh,  4ft.  thick  :  in  what  time  could  20  men  build  a  wall 
87 rd.  long,  8ft.  high,  and  5ft.  thick?         Ans.  58;;;; da. 


PRACTICE.  209 

11.  If  180  men,  in  6  da.  of  10  lir.  each,  dig  a  trench 
200  yd.  long,  3yd. wide,  2yd.  deep;  in  how  many  days 
can  100  men,  working  8hr.  a  day,  dig  a  trench  180  yd. 
long,  4  yd.  wide,  and  3  yd.  deep  ?  Ans.  24.3  da, 

J|^*  For  additional  problems,  see  Ray's  Test  Examples. 


XV.   ALIQUOTS,   OR   PRACTICE. 

Art.  206.  One  number  is  an  aliquot  part  of  another, 
when  it  will  exactly  divide  it  (Art.  110).  Thus,  5  cents, 
10  cts.,  20  cts.,  &c.,  are  aliquot  parts  of  $1. 

TABLE   OF   ALIQUOT   PARTS   OF   A  DOLLAR. 
100    cents  =  .     .a  dollar. 


50    cents  =  -^  of  a  dollar. 

25    cents  =  |  of  a  dollar. 

12^  cents  =  ^  of  a  dollar. 

6J  cents  =j^g  of  a  dollar. 


20    cents  =  i  of  a  dollar. 
1 0    cents  =  J^  of  a  dollar. 


5  cents  =  5^^  of  a  dollar. 
33 1  cents  =  |  of  a  dollar. 
1G|  cents  =  J  of  a  dollar. 


CASE    I. 

Art.  207.  To  find  the  cost  of  articles,  when  the  price  is 
an  aliquot  part,  or  aliquot  parts  of  a  dollar. 

1.  What  cost  24  yards  of  muslin,  at  25  cts.  a  yd.? 

Solution. — If  the  price  was  $1  a  yard,  the  cost  would  be  $24; 
and,  since  25  cents  is  1  of  a  dollar,  the  cost  at  25  cts.  a  yard  will 
be  i  the  cost  at  $1 ;     and  1  of  $24 =$6.     Ans. 

2.  What  cost  16  yards  of  calico,  at  374  cts.  a  yard  ? 

Solution. — If  the  price  was  $1  a  yard,  the  cost  would  be  $16„ 
At  25 cts.  a  yard,  the  cost  would  be  1  the  cost  at  $1 ;  i  of  $16==$4. 

Again;  since  12^ct8.  is  ^  of  25cts.,  the  cost  at  12Jcts.  a  yard, 
will  be  ^  the  cost  at  25 cts. ;  1  of  $4  =$2. 

But,  the  cost  at  37Jcts.  a  yard,  is  equal  to  the  sum  of  the  costs 
at  25 cts.  and  at  12^  cts.;  $4-|-$2  =  $6.  Ans. 

Review. — 206.  When  is  one  number  an  aliquot  part?    Give  examples. 
3d  Bk.  14 


210 


RAY'S   PRACTICAL    ARITHMETIC. 


*3.  What  cost  24  yd.  silk  at  62^  cts.  a  yd.?  Am.  $15. 

Rule  for  CaS3  I. —  Take    such    aliquot  parts   of  the   cost 
at  $1,  as  may  be  necessary  tojind  the  cost  at  the  given  price. 

4.  What  cost  48yd.  of  linen,  at  68|  cts.  per  yd.? 

848  =  cost  of  48  yards,  at  31. 


OPERATION. 


50    cts.  =  i 

?A- 

U 

12i  cts.  —  1 

6  = 

(( 

6]  cts.  —  i 

Rrr. 

U 

Ans.  $33  = 


a 


a 


(( 


a 


u 


a 
a 

at  50    cts. 
at  V2\  cts. 
at    6|  cts. 

u 

at  68|  cts. 

What  will  be  the  cost 

5.  Of  173  bu.  corn,  at  25  cts.  a  bu.?     Am.    $43.25 

6.  45  lb.  cheese,  at  31i  cts.  a  lb.?  Am.  $14.06] 

7.  54  yd.  calico,  at  43|  cts.  a  yd.?  Ans.  $23.62^ 

8.  32  bu.  rye,  at  93|  cts.  a  bu.?      Am.    $30.00 

9.  What  cost  20yd.  cloth,  at  $3.12A  per  yd.? 


OPERATION.   $20  =  cost  of  20  yd.  at  $1. 
3      


The  cost  at  $1  a 
yard  is  multiplied 
oy  the  number  of 
dollars  (3);  and 
for  the  12;\  cts.,  an 
aliquot  part,  (J)  of  ^ws.  $62.50=  cost  at  $3.12^ 

the  cost  at  $1  is  taken  j  the  results  are  added. 


12  J  cts.  =  1 


60       =costat$3.00 
2. 50=  cost  at      .121 


Find  the  cost  of 

10.  80  gal.  of  wine,  at  $2.37A  a  gal. 

11.  36  bl.    of  flour,  at  $8.87^  a  bl. 

12.  77  gal.  of  wine,  at  $1.62^-  a  gal. 

13.  175  A.    of  land,  at $14. 37A  an  A. 

14.  224  bl.    of  flour,  at  $3.43|  a  bl. 

15.  462  yd.  of  cloth,  at  $1.06]  a  yd. 

16.  185  yd.  of  cloth,  at  $1.33|ayd. 

17.  150  yd.  of  satin,  at  $3. 66|  a  yd. 

18.  24  yd.  of  silk,    at  $1.16|ayd. 


ANSWERS. 

$190.00 

$319.50 

$125. 12i 

$2515.621 
$770.00 

$490.87;V 

$246. 66 f 

$550.00 

$28.00 


PRACTICE.  211 

Art.  208.  Case  II. —  To  find  the  cost  of  a  quantity  con' 
sisting  of  several  denominations. 

1.  What  will  be  the  cost  of  4 A.  IK.  20 R  of  land, 
at  ^16.40  per  A.? 

OPERATION.        ^16.40  =  cost   of      1  A. 

4      


lR.=  i  A. 

20P.=  i  R. 


65.60  =  cost  of  4  A. 
4.10  =  cost  of  1  R. 
2.05  =  cost  of  20  P. 


Ans.  $71.75  =  cost  of   4  A.  1  R.  20  P. 

Kule  for  Case  II. — Multiply  the  price  hy  the  vvmber  of 
the  denomination  at  which  the  price  is  rated,  and  find  the  cost 
of  the  lower  denominations,  by  taking  aliquot  parts.  Add  the 
different  costs ;  the  sum  will  be  the  cost  of  the  whole. 

Find  the  cost  of  answers. 

2.  14  A.  2 R.,  at  SIO. 00  per  A.  $145.00 

3.  28  A.  3  R.,  at  §12.50  per  A.  S359.37| 

4.  5  A.  1  R.  10  P.,  at  $12  per  A.  $63.75 

5.  12  A.  1  R.  10  P.,  at  $18  per  A.  $221. 62i 

6.  14  A.  3  R.  25  P.,  at  $12.50  per  A.       $186.32i| 

7.  3  yd.  2  qr.,  at  $1.75  per  yd.  $6.12^ 

8.  4  yd.  3  qr.,  at  $1.50  per  yd.  $7.12| 

9.  56yd.3qr.,  at  $17.25  per  yd.  $978.93} 

10.  83  bu.  3  pk.  2  qt.,  at  $6  a  bu.  $502.87^ 

11.  24bu.3pk.  7qt.,  at  $4  a  bu.  $99. 87^ 

12.  40bu.3pk.  7qt.  1  pt.,  at  $3.20  a  bu.  $131.15 

13.  17  bu.  1  pk.  1  qt.  1  pt.,  at  $2.56  a  bu.  $44.28 

14.  5  lb.  11  oz.  butter,  at  24  cts.  per  lb.  $1.36^ 

15.  3  lb.  13  oz.  15  dr.  spice,  at  $2 .  56  a  lb.       $9 .  91 

16.  17  A.  3 R.  39  P.,  at  $3.20  per  A.  $57.58 

B^""  For  additional  problems,  see  Rays  Test  Examples. 


Review. — 207.    What    is    Case  1  ?      What    the    Rule    for    Case   1  ? 
208.  What  is  Case  2  ?    What  the  Rule  for  Case  2  ? 


212  RAY'S   PRACTICAL   ARITHMETIC. 


XVI.    PERCENTAGE. 

Art.   209.  Percentage  embraces  calculations  in  which 
reference  is  made  to  a  hundred  as  the  unit ;  and, 

The  per  cent,  or  percentage  of  any  number  or  quantity, 
is  so  many  hundredths  of  it.     Thus, 

1  per  cent,  is  jj^  ;     2  per  cent.  y§^ ;     3  per  cent.  j|^. 

Since   jj^j  =  .01,    y^^  =  .02,    and  so   on,    (Art.  176), 
therefore,  per  cent,  may  be  expressed  decimally  ;  thus. 


4  per  cent,  is  .04 

5  per  cent,  is  .05 
9  per  cent,  is  .09 


10  per  cent,  is  .10 

25  per  cent,  is  .25 

75  per  cent,  is  .75 


100  per  cent,  (jgg,  or  the  whole,)  is 1.00 

125  per  cent.  (j§§,  or  -fg^  more  than  the  whole,)  is  1.25 
^  per  cent,  is  a  half  of  1  per  cent.,=^  of  yj^^,  is  .005 
^  per  cent,  is  i  of  1  per  cent., = J  of  y^^,  is  .00125 

3^  per  cent,  is  .03|,=.035  is .035 

EXAMPLE. — Write  7  per  cent,   decimally.     4A   per  ct. 
5|  per  ct.     10^  per  ct.     37A  per  ct.     120  per  ct. 

J8^°'The  Sign  ^,  is  used  in  business,  instead  of  the  words 
"  per  cent.  ;"   thus,  5  per  cent,  is  written  5  %. 

Art.  210.  Case  I. —  To  find  any  per  cent,  of  a  number. 
1.  What  is  2  %  (2  per  cent.)  of  $125?       ^p^^,.,,^^. 

Sol. — To  take  2  per  cent,  is  to  take  2  bun-  $125 

dredths.     Hence,   multiply  by  2  and  divide  by  •  02 

100;  or  multiply  by  yS,  expressed  decimally.         .      ~9~^(7 

*2.  What  is  7  %  of  175  dollars  ?  Ans.  $12.25 

Review. — 209.  What  does  Percentage  embrace  ?    What  is  the  per  cent, 
of  any  number  ?     Give  examples. 


PERCENTAGE. 


213 


Rule  for  Percentage. — Multiphj  hy  tJie  rate  per  cent,  ex- 
pressed decimally;  the  product  will  he  the  per  cent,  required. 

Or,  by  Proportion  (Art.  203).  ^5  100  is  to  the  rate  per  cent, 
so  is  the  given  number  or  quantity  to  the  required  per  cent. 


3.  What  is  51  %  of  $150? 

Suggestion.  —  When  the  rate 
per  cent,  is  a  common  fraction, 
or  mixed  number,  multiply  as 
explained  in  Art.  152 ;  and  in 
pointing  the  product,  count  only 
two  decimals  in  the  multiplier. 


OPERATION. 

$150 

7.50  =  5     percent. 
.50=    ^  per  cent. 


Am.  $8.00  =  51  percent. 


ANSWERS. 

4. 

What  is 

6    %  of  $250  ? 

$15.00 

5. 

What  is 

7    9^  of  $162? 

$11.34 

6. 

What  is 

5    %  of  $118? 

$5.90 

7. 

What  is 

8    %  of  $11? 

$0.88 

8. 

What  is 

1    %  of  $278  ? 

$2.78 

9. 

What  is 

2 1  %  of  $68  ? 

$1.53 

10. 

What  is 

41  %  of  $220.50? 

$9.9224- 

11. 

What  is 

71  %  of  $115.42? 

$8,656+ 

12. 

What  is 

53  %  of  $243.16? 

$13.981-f 

13. 

What  is 

31  %  of  $1250? 

$40 . 625 

14. 

What  is 

25    %  of  $25  ? 

$6.25 

15. 

What  is 

lOU  %  of  $2002? 

$2032.03 

16. 

What  is 

208^  %  of  $650? 

$1352.00 

17. 

What  is  : 

LOOO    %  of  $24.75? 

$247.50 

18. 

What  is 

^\  %  of  $400  ? 

$0.40 

19. 

What  is 

1    %  of  $464  ? 

$1.74 

20. 

What  is 

J3  %  of  $1950? 

$1.62| 

Review. — 209.  How  may  per  cent,  be  expressed  ?  Give  examples. 
What  Is  100  per  cent.  ?  What  is  125  per  cent.  ?  ^  per  cent.  ?  3A  per 
cent.  ?    What  sign  is  used  for  per  cent.  ?    Give  examples. 


214  RAY'S    PRACTICAL    ARITHMETIC. 

21.  If  8^  %  of  $72  be  taken  from  it,  how  much  will 
remain?  Am.  865.88 

22.  I  had  S800  in  bank,  and  drew  out  36  %  of  it : 
how  much  had  I  left?  Am.  8512. 

23.  "What  is  the  difference  between  131  ^  of  856,  and 
14^  %  of  851?  Am.  13cts. 

24.  A  merchant  in  expending  81764,  paid  23  ^  for 
cloth ;  31  f^Q  for  calicoes  ;  9  ^  for  linens  ;  3i  ^  for 
silks  ;  the  remainder  for  muslin  :  how  much  did  he  paj 
for  each  ?  Am.  to  last,  8595.35 

25.  A  grocer  bought  4  bags  of  coffee,  of  75  lb.  each : 
12^  %  was  lost  by  waste  :  what  was  the  rest  worth,  at 
14cts.  per  lb.?  Am.  836.82 

26.  A  flock  of  160  sheep  increased  35  ^  in  1  year: 
how  large  was  it  then?  Am,  216. 

27.  I  had  320  sheep  ;  5  %  were  killed  :  after  selling 
25  ^  of  the  rest,  how  many  were  left  ?  Ans.  228. 

28.  A's  salary  is  8800  a  year  :  he  spends  18  %  for 
rent ;  15  ^  for  clothing  ;  23  %  for  provisions  ;  12  %  for 
sundries  :  how  much  is  left?  Am.  8256. 

Art.  211.   case   ii. 
To  find  what  per  cent,  one  numher  is  of  another. 

1.  What  %  of  8  dollars  is  82?  operation. 

Solution.— One    per    cent,    of  $8  .08)2.00 

is$8x.01=.08,  (Art.210)i  and  since  Am.  25  percent 

$2  contain  this  25  times,  it  must  bo 
25  per  cent,  of  $8.     Or  thus:    the    o_,_    or    ^9- 
question  is  the  same  as  to  find  how    5  — 4  =  '-5  =  ^5  percent 
many  hundredths  $2  are  of  88.     Now  2  is  2  of  8,  and  |='=.25=: 
25  per  cent 

*2.  What^of815,  is83?  Am.  20. 


Review.— 210.  How  is  the  percentage  of  .a  number  found,  Ruk  ? 
How,  when  the  nito  per  cent,  is  a  mixed  number  or  a  fraction  ?  211.  What 
is  Case  2  ?     Wliat  the  Rule  for  Case  2  ? 


I 


COMMISSION.  215 

Rule  for  Case  II. — Divide  the  number  which  is  considered 
the  percentage^  by  I  per  cent,  of  the  other  number;  the  quotient 
will  be  the  required  rate  per  cent. 

Or,  by  Proportion. — As  the  given  number  is  to  the  percentage, 
so  is  100  to  the  required  rate  per  cent. 

3.  What  %  of  S50,  is  $6  ?  Ans.  12. 

4.  What  %  of  ^75,  is  $4.50?  Ans.  6. 

5.  What  %  of  $9,  is  $3?  '     Ans.  33 i. 

6.  What  %  of  $25,  is  $0.25?  Ans.  1. 

7.  What  %  of  $142.60,  is  $7.13?  Ans.  5. 

8.  What  %  of  $9,  is  $9  ?  Ans.  100. 

9.  What  %  of  $9,  is  $13.50  ?  ^Ins.  150. 

10.  What  %  of  $243,  is  $8,505?         Ans.  3.5==3i. 

11.  What  %  of  $2,  is  2  mills?  Ans.  Jg. 

12.  What  %  of  $3532,  is  $13,245?  Ans.  |. 

13.  A  man  had  $300,  and  spent  $25 :  what  %  was 
that  of  the  whole?  Ans.  8|. 

14.  A  man  owed  $500,  and  paid  $75  of  it :  what  %  of 
the  whole  debt  did  he  pay  ?  Ans.  15. 

15.  What  %  of  any  number  is  |  of  it?  Ans.  60. 

16.  A  miller  takes  for  toll,  6qt.  from  every  5bu.  of 
grain  ground:  what  5^  does  he  get?  Ans.  3|. 

g^°  For  additional  problems,  see  Ray's  Test  Examples, 

APPLICATIONS    OF    PERCENTAGE. 

Art,  212.  The  principal  applications  of  percentage  are 
^n  Commission,  Insurance,  Stocks,  Brokerage,  Interest, 
Discount,  Profit  and  Loss,  Duties,  and  Taxes. 

Art.  213.  COMMISSION 

Is  the  sum  allowed  to  Agents  for  buying;,  selling,  or 
transactino:  other  business. 

Those  who  transact  business  for  others,  are  Agents, 
Commission  Merchants,  Factors,  or  Correspondents. 


216  RAY'S  PRACTICAL  ARITHMETIC. 

Commission  is  estimated  at  a  certain  rate  jyer  cent,  on 
the  amount  employed  in  the  transaction ;  it  is  calculated 
by  the  Rule,  Art.  210. 

1.  At  5^,  what  does  an  agent  receive  for  selling 
goods  amounting  to  S240?  Ans.  S12.00 

2.  At  2i  %,  what  is  the  commission  for  selling  goods 
amounting  to  S460?  Ans.  §11.50 

3.  What  the  commission  on  sales  of  S180  at  4^, 
and  on  sales  of  8119  at  3  %  ?  Ans.  $10.77 

4.  What  the  commission  on  $240  sales  at  3  ^,  and  on 
8225  sales  at  5  %  ?  Ans.  S18.45 

5.  What  the  commission  on  sales  of  ^80  at  2|  % ; 
8275  at  3i  %  ;  and  $216  at  2^  %  ?  Ans.  $25.36 

6.  What  on  sales  of  $275  at  3  %  ;  $341  at  15%; 
$964  at  25  %  ;  and  $217  at  2^  %  ?        Ans.  $305.2825 

7.  An  agent  sells  25  bl.  of  molasses,  at  $13  each,  at 
2^  %  commission :  what  will  the  agent  and  owner  each 
receive?  -4ns.  Agent,  $8.12 A  ;  Owner,  $316.87^ 

8.  A  factor  sells  15hhd.  sugar,  of  11141b.  each,  at 
Sets,  a  lb.,  charging  3|  %  commission:  what  sum  will  he 
pay  the  owner?  Ans.    $1293.354 

9.  A  factor  sells  250 bl.  pork,  at  $15  each;  175 bl. 
flour,  at  $7  each;  and  14561b.  feathers,  at  25cts.  a  lb.; 
his  commission  is  3  %  :  what  sum  will  he  pay  the  owner? 

Ans.  $5178.83 

Art.  214.  When  an  agent  invests  money,  his  com- 
mission is  computed  on  the  amount  he  expends. 

10.  An  agent  receives  $210  to  buy  goods;  after  deduct- 
ing his  commission  of  5  %,  what  sum  must  he  expend? 

Solution, — For  each  $1  expended,  he  receives  5ctg.  Hence,  for 
each  $1.05,  received,  he  must  expend  $1 ;  therefore,  he  must  expend 
as  many  dollars  as  $1.05  are  contained  times  in  $210: 

That  is,— $210  -h$  1 .05  =  200.     Ans.  $200. 
Peoof.— $200X.05  =  $10;  and  $200 -|- $10  =  $210. 


INSURANCE.  217 

Hence,  divide  the  given  sum  by  $1,  increased  by  the  com- 
mission on  SI ;  the  quotient  loill  be  the  sum  to  be  expended;  the 
difference  between  this  and  the  given  sum  is  the  commission. 

11.  A  received  ^312  to  buy  goods:  after  deducting  "his 
commissions  at  4  ^ ,  what  must  he  expend?     Ans.  $300. 

12.  B  receives  $1323.54  to  buy  goods,  at  8  %  com- 
mission: what  was  his  commission?  Ans.  $98.04 


Art.  215.  insurance 

Is  security  against  loss  by  fire,  navigation,  &c.  It  is 
effected  with  a  company,  or  an  individual,  who,  for  a 
certain  per  cent.,  agrees  to  make  good  the  amount  in- 
sured, for  a  certain  period. 

The  Policy  is  the  written  contract  between  the  parties. 

The  Premium  is  the  sum  paid  for  insurance. 

The  Insurer  is  called  an  Underwriter. 

Insurance  on  ships,  boats,  &c.,  is  Marine  Insurance. 

The  Rate  of  insuring  is  a  certain  per  cent,  of  the  amount 
insured,  and  is  calculated  by  Rule,  Art.  210. 

1.  What  is  paid  for  insuring  $2250  on  a  house,  at 
1^%  premium,  the  policy  costing  $1?         Ans.  $34.75 

2.  A  steamboat  is  valued  at  $12600,  the  cargo  at 
$14400  ;  I  of  their  value  is  insured  at  4^  ^ ,  the  policy 
costing  $1:  what  the  cost  of  insurance?       Ans.  $811. 

3.  What  is  paid  for  insuring  |  of  a  house  worth  $5000, 
at  1  %,  the  policy  costing  $1.50?  Ans.  $20.25 

4.  A  has  a  ship  valued  at  $21000,  and  a  house  at 
$1200:  what  will  be  the  cost  of  insuring  ^  of  the  ship  at 
12^%,  and  the  full  value  of  the  house,  at  1^%,  the 
two  policies  costing  $1  each?  Ans.  $1520. 

Eeview. — 212.  Where  are  the  principal  applications  of  percentage? 
213.  What  is  commission  ?    How  estimated  ? 

215.  What  is  insurance?  How  effected?  What  is  the_policy?  The 
premium  ?   An  underwriter  ?   Marine  insurance  ?   Eate  ?   How  calculated  ? 


218  RAY'S    PRACTICAL   ARITHMETIC. 

Art.  216.  Some  persons  wish  to  insure  so  as  to  cover 
the  property  and  premium  paid,  in  case  of  a  loss. 

5.  A's  shop  is  valued  at  8190  :  at  5  ^,  what  sum  must 
be  insured,  so  that  he  may  recover  the  value  of  the 
property  and  the  premium  in  case  of  loss  ? 

Solution. — Since  the  rate  of  insurance  is  5  per  cent.,  5  cents 
of  every  dollar  insured  is  premium,  and  the  remaining  95  cents 
property ;  hence, 

As  often  as  the  property  to  be  insured  contains  95  cts.,  so  many 
dollars  must  be  insured  to  cover  that  property  and  its  premium. 

But,  $190-^.95  =  8200.  Ans. 

Proof.— 8200  X.  05  =  810;   and  8190  +  810  =  8200. 

In  such  cases,  divide  the  value  of  the  property  insured^  by  the 
remainder  left  after  subtracting  the  rate  of  insurance  from  1. 

6.  At  1  %,  what  sum  must  be  insured  on  82475,  to 
secure  both  property  and  premium  ?  Ans.  82500, 

7.  At  12 J  %,  what  sum  must  be  insured  on  813125, 
to  cover  property  and  premium?  Ans.  815000. 

8.  At  IJ  %  what  sum  must  be  insured  on  82358,  to 
cover  the  premium  and  property  ?  Ans.  82400. 

9.  At  I  %  what  must  be  paid  for  insuring  82287.39, 
to  cover  both  property  and  premium?  Ans.  88.61 

Art.  217.   stocks. 

Stock  is  money  or  property  invested  in  Manufactories, 
Railroads,  Insurance  Companies,  Banks,  Bonds,  &c. 

Stock  is  divided  into  shares,  usually  of  850  or  8100 
each.     The  owners  are  Stockholders. 

The  original  cost  of  a  share  is  its  pa?*  or  nominal  value. 

When  8100  of  stock  sells  for  8100,  the  stock  is  said 
to  be  at  par ;  when  it  sells  for  more  than  8100,  above  par  ; 
and  when  for  less  than  8100,  below  ptar. 

Review. — 216.  IIow  find  the  sum  to  bo  insured,  to  cover  both  prop- 
erty and  premium? 
217.  What  id  stock?    How  divided  ?    What  is  meant  by  par  value  ? 


STOCKS.— BROKERAGE.  219 

When  stocks  are  above  par  they  are  at  an  advance;  and 
when  below  par ^  at  a  discount. 

The^er  cent,  is  calculated  on  the  par  value.     Art.  210. 

1.  What  is  the   value   of  $1400   of  stock,  at   104  %  ; 
that  is,  4  %  above  par  ?  Ana.  $1456. 

2.  What  the  value  of  $1400  of  bank  stock,  at  96  %  ; 
or  4  %  below  par?  Ans.  $1344. 

3.  What   the  value  of  11    shares  railroad   stock  ($50 
each)  at  5  %  advance?  Ans.  $577.50 

4.  What  the  value  of  15  shares  canal  stock  ($75  each), 
at  10  %  below  par?  Aiis.  $1012.50 

5.  Bought  $1500  of  bank  stock,  at  104i  %,  and  sold 
it  at  1081  %  :  find  the  gain.  Ans.  $57.50 

6.  Bought  $1300  of  stock,  at  3  %  below  par,  and  sold 
it  at  5i  %  above  par  :  find  the  gain.  Ans.  $110.50 

7.  Bought  $860  bank  stock,  at  4  %  advance ;  sold  at  a 
discount  of  2i  %  :  find  the  loss.  Ans.  $55 .  90 


Art.  218.  BROKERAGE. 

Dealers   in    money,   stocks,    &c.,  are    Brokers.      The 
operation  of  finding  the  percentage  is  termed  Brokerage. 

The    percentage    is    calculated.   Art.   210,   on    the   par 
value  of  the  funds  received  or  paid. 

When  bonds,  notes,  or  coin,  bring  more  than  their  nominal  value, 
they  are  at  a  premium ;  when  less^  at  a  discoutU. 


1.  Find  the  premium,  at  IJ  %,  on  $600.         Ans. 

2.  What  sum  is  lost  in  exchanging  $289  in  bank  notes, 
at  a  discount  of  1|  %  ?  Aiis.  $3.61i 

3.  I  sold  $360,  in  bank  notes,  at  |  ^  discount :  how 
much  did  I  receive  ?  Ans.  $358 .  65 

Review. — 217.    "When   is   stock   at   par  ?     Above   par  ?     Below   par  ? 
At  an  advance  ?    At  a  discount  ?    On  what  is  the  per  cent,  calculated  ? 


220  RAY'S   PRACTICAL    ARITHMETIC. 

4.  What  sum  must  a  broker  pay  for  SI  34,  in  bank 
notes,  at  2^  %  discount?  An&.  S130.65 

5.  What  sum  must  a  broker  pay  for  $200  in  gold, 
at  I  %   premium?  '  $201.25 

6.  A  buys  S500,  in  notes,  at  i  ^  discount,  and  sells 
at  I  %  premium:  find  his  gain.  An&.  $3.75 

Art.  219.  INTEEEST. 

Interest  is  an  allowance  made  by  the  borrower  to  the 
lender,  for  the  use  of  money. 

Principal  is  the  sum  loaned,  for  which  interest  is  paid. 

Amount  is  the  sum  of  the  principal  and  interest. 

Per  annum  signifies  for  one  year. 

Rate  per  cent,  per  annum^  is  the  number  of  dollars 
paid  for  the  use  of  $100,  or  the  number  of  cents  paid  foi 
the  use  of  100  cents,  for  one  year. 

Illustration. — B  borrowed  of  A  $200  for  1  year,  and  agreed 
to  pay  him  $12  for  the  use  of  it,  tliat  is,  $6  for  the  use  of  $100  foj 
one  year.  Here,  $200  is  the  principal,  $12  the  interest,  $6  the  rate 
per  cent.,  and  $212  the  amount. 

Rem. — 1.  The  distinction  between  Interest  and  other  computa- 
tions in  Percentage,  as  Commission,  Brokerage,  «&c.,  is  this :  in 
the  latter,  the  rate  per  cent,  has  no  regard  to  time,  whereas, 

In  Interest,  time  is  always  considered  :  and. 

For  periods  of  time  greater  or  less  than  one  year,  the  interest  is 
proportionally  greater  or  less  than  the  interest  for  one  year. 

2.  For  brevity,  the  termjoer  annum  is  seldom  used  in  connection 
with  rate  per  cent.,  but  is  always  understood.     Hence, 

In  Interest,  rate  per  cent,  always  means  rate  per  cent,  per  annum. 

Legal  Interest  is  the  rate  established  by  law.  Any 
rate  higher  than  the  legal  rate  is  Usury. 

Review. — 218.  What  arc  brokers?  "What  is  brokerago  ?  On  what  is 
the  percentage  calculated  ?   "When  arc  notes  at  premium  ?   At  discount? 

219.  "What  is  Interest ?  Principal?  Amount?  "What  does  per  annum 
signify  ?    "What  the  rate  per  cent,  per  annum  ? 


SIMPLE    INTEREST.  221 

When  no  rate  per  cent,  is  named,  the  legal  rate  where  the 
business  is  transacted,  is  understood. 

Note. — The  legal  rate  of  Int.  in  Louisiana  is  5  per  cent.; 

InN.  York,  S.Carolina,  Michigan,  Wisconsin,  and  Iowa,  7  per  cent.; 
In  Georgia,  Alabama,  Mississippi,  Florida,  and  Texas,  8  per  cent.; 
In  the  rest  of  the  States,  and  in  the  U.  S.  Courts,  6  per  cent. 

CASE    I. 

Art.  220.  To  find  the  interest  on  any  principal,  at  any 
rate  per  cent,  for  one  or  more  years. 

1.  Find  the  interest  of  $25  for  1  yr.,  at  6  %. 

SoLUTiox.— Since    6    per   cent,    is   -6-=. 06,  ^^^^'^gp?' 

it  is  only  necessary,  as  in  Percentage,  Art.  210,  ^^ 

to  multiply  the  principal  by  the  rate  expressed  * 

decimally,  and  the  product  will  be  the  Int.  for  1  yr.  Ans.  §1 .  50 

Or,  thus:  Six  per  cent,  is  $6  on  $100,  or  6  cents  on  $1.  Hence, 
if  the  interest  on  $1  is  6  cts.=$.06,  the  interest  on  $25,  for  the 
same  time,  will  be  25  times  as  much;  25  X$- 06  =  $1.50  Ans. 

2.  Find  the  interest  of  $50  for  3  yr.,  at  7  %. 

rr,,  -r  «  OPERATION. 

Solution. — The  Int.  of  any  <Q^r. 

Bum   for  3  yr.   is,   3   times  as  r^-r 

much  as  its  Int.  for  1  yr.  ! 

Find  the  Int.  of  $50  for  1  yr.,  $3 .  50  =  Int.  for  1  yr. 

which  is  $3.50;  multiply  this  ^ 

by  8  to  obtain  the  Int.  for  3  yr.  ^^^   $10.50  =  Int.  for  3  yr. 

*3.  Interest  of  $65  for  4  yr.,  at  5  %.  Ans.  $13. 

Rule  for  Case  I. — 1.  Multiply  the  principal  by  the  rate 
per  cent,  expressed  decimally ;  the  product  will  be  the  interest 
for  one  year. 

Eeview. — 219.  Eem.  1.  What  is  the  distinction  between  interest  and 
commission  ?    2.  What  does  rate  per  cent,  in  interest  always  mean  ? 

219.  What  is  legal  interest  ?  What  is  usury  ?  When  no  rate  is  named, 
what  is  understood  ?  Note.  In  what  State  is  the  legal  rate  of  interest  5  per 
cent.  ?    In  what  7  per  cent.  ?    In  what  8  per  cent.  ?    In  what  6  per  cent.  ? 


222  RAY'S   PRACTICAL   ARITHMETIC. 

2.  Multiply  the  interest  for  one  year  by  the  given  number  of 
years  ;  the  product  will  be  the  Int.  required. 

3.  To  find  the  amount,  add  the  principal  to  the  interest 

Required  the  Interest  answers. 

4.  Of  $200.00  for    1  year,    at  8  %.  $  16.00 

5.  $150.00  for    1  year,    at  5  %.  $    7.50 

6.  $300.00  for    2  years,  at  6  %.  $  36.00 

7.  $275.00  for    3  years,  at  6  %.  $  49.50 

8.  $187.50  for    4  years,  at  5  %.  $  37.50 

9.  $233.80  for  10  years,  at  6  %.  $140.28 

10.  Find  the  amount  of  $225.18  for  3yr.,  at  ^  %. 

225.18  OPERATION. 


.041 


Suggestion.  —  When  the  rate 
contains  a  fraction,  as  in  this  ex- 
ample, write  it  as  .04^  or  .045  90072  =  Int.  at  4  %. 

When  the  fraction   can  not  be  1125  9=Int..  at  ^  fo. 

exactly  expressed  in  decimals,  ex-       -in    i  o  o  i       t       /.     i 

."l  ^      ..  lU.iod  l  =  Int.  for  1  yr. 

press  It  as  a  common  fraction.  o 

In  adding  the   principal  to  the     

Int.,  place  figures  of  the  same  order      30.3993=Int.  for  3  yr. 
under  each  other.  225.18       =principal. 

$255.57  93  =  amount. 
"What  is  the  Amount  of 

ANSWERS. 

11.  $215.00  for  1  year,    at  6  %  ?  $227.90 

12.  $  45.00  for  2  years,  at  8  %?  $52.20 

13.  $  80.00  for  4  years,  at  7  %  ?  $102.40 

14.  $420.00  for  1  year,    at  5j  %  ?  $442.40 

15.  $237.16  for  2  years,  at  3|  %  ?  $254,947 

16.  $  74.75  for  5  years,  at  4  %  ?  $  89.70 

17.  $  85.45  for  4  years,  at  0  %  ?  $105,958 

Review, — 220.  What  is  Case  1  ?    How  find  the  interest  on  any  prin- 
cipal, at  any  rate,  for  one  or  mure  years.  Rule  for  Case  1  ? 


SniPLE   INTEREST.  223 


18.  8325.00  for  3  years,  at  5|  %  ?      Ans.  S377.65 

19.  $129.36  for  4  years,  at  4?  %  ?      Ans.  $151,998 

CASE    II. 

Art.  221.  To  find  the  interest  of  any  principal^  at  avy 
rate  per  cent. ^  for  any  number  of  months^  or  days. 

1.  Find  the  interest  of  $300  for  1  mon.,  at  6  ^. 

Solution. — Since  1  mon.  is  X  $300     operation. 

of  a  yr.,  the  Int.  for  1  mon.  is  ^  .06 

of  the  Int.  for  a  yr.     But  the  Int. 

of  $300  for  1  yr.  is  $18.00,  and  J^  ^  ^)1  H.U0=:Int.  for  1  yr. 

of  this  is  $1.50     Ans.  $1 . 5  0=lnt.  for  1  mon. 

*2.  The  Int.  of  $240  for  2  mon.,  at  8  % .    Ans.  $3 .  20 

On  this  principle,  the  Int.  for  3  mon.  is  -,^o  =^  t  ^^^^  ^^^-  ^or  a  jr. ; 
for  4  mon.  it  is  -%  =  1 ;   for  5  mon.  -%,  &c. 

3.  Find  the  interest  of  $288  for  1  day,  at  5  % . 

o  AiJ-ii>  $288        OPERATION. 

Solution. — As   1   da.    is  J^  of  ak 

a  mon.,  the  Int.  for  1  da.  is  ^L  of  ' 

the    Int.    for    1    mon.     The    Int.      1 2)1 4.4 0=lnt.  for  1  year. 

for  1  mon.  is  $1.20,  and  J^j  of  this         ^TTT;    TTT 

.    cic^^      A  ^^  30)1.2  0=Int.  fori  mon. 

$.04=:Int.  for  1  day. 
*4.  The  Int.  of  $360  for  2  days,  at  6  % .     Ans.  $0 .  12 

On  the  same  principle,  the  interest  for  3  da.  is  _3____i_  of  the  Int. 
for  1  mon.;  for  4  da.  it  is  ^Q=-f^]  for  5  da.,  /o=Jm  for  6  da.; 
it  is  365-1;  for  7  da.,  J^. 

Rule  for  Case  II. — Find  the  interest  of  the  given  sum  for 
one  year^  (Art.  210). 

To  find  the  Int.  for  any  number  of  months^  take  such  a  part 
of  this.,  as  the  given  number  of  months  is  part  of  a  year. 

To  find  the  Int.  for  any  number  of  days.,  take  such  apart 
of  the  Int.  for  1  mon..,  as  the  days  are  part  of  a  month. 


224 


RAYS    PRACTICAL    ARITHMETIC. 


5.  Find  the  interest  of  S50  for  5  mon.,  at  6  %. 


Suggestion. — After  find- 
ing the  interest  for  1  year, 
find  the  Int.  for  months, 
by  taking  aliquot  parts 
(Art.  208) ;  or, 

Multiply  by  the  number 
of  mon.,  and  divide  by  12, 
(Art.  152). 


OPERATION. 

.06 


4  mon.=  i 
1  mon.=^ 


3.00=Int.  lyr. 


1 .  00  =  Int.  for  4  mon. 
.25  =  Int.  for  1  mon. 


^1  o  25  =  Int.  for  5  mon. 
To  find  the  Int.  for  da.,  take  aliquot  parts  of  the  Int.  for  1  mon.; 


or,  multiply  by  the  number  of  da.,  and  divide  by  30. 

Find  the  Interest  of 

6.  $86.00  for    3  mon.,  at  6  %. 

7.  $50.00  for    4  mon.,  at8  %. 

8.  $150.25  for    6  mon.,  at  8%. 

9.  $360.00  for    7  mon.,  at5  %. 

10.  $204.00  for  11  mon.,  at  7  %. 

11.  $726.00  for  10  days,  at  6  %. 

12.  $1200.00  for  15  days,  at  6  ^. 

13.  $180.00  for  19  days,  at  8  %. 

14.  $240.00  for  27  days,  at  7  %. 

15.  $100.80  for  28  days,  at  5  %. 


Find  the  Amount  of 

16.  $228.00  for    9  mon.,  at  6  %. 

17.  $137.50  for    8  mon.,  at  6  %. 

18.  $150.00  for  18  days,  at  5  %. 

19.  $360.00  for  11  days,  at  6  %. 

20.  $264.00  for    9  days,  at  6  %. 
fii^*  For  additional  problems,  see  Ray's  Test 


ANSWERS. 

.  $1.29 
.  $1.33| 
.  $6.01 
.  $10.50 
.  $13.09 
.  $1.21 
.  $3.00 
.  $0.76 
.  $1.26 
.  $0,392 

ANSWERS. 

$238.26 
$143.00 
$150,375 
$360.66 
$264,396 
Examples. 


Ekview. — 221.  How  find  tho  interest  of  any  sum  for  any  number  uf 
months,  Rule  for  Case  II  ?    For  any  number  of  days  ? 


SIMPLE    INTEREST. 


225 


Art.   222.    GENERAL   RULE    FOR    INTEREST. 

1.  For  one  year.  Multiply  the  Principal  by  the  rate  per  cent, 
expressed  decimally. 

2.  For  more  years  than  one.  Multiply  the  Interest  for  1  year 
by  the  number  of  years. 

3.  For  months.  Take  such  part  of  the  Interest  for  1  year,  as 
the  number  of  months  is  part  of  one  year. 

4.  For  days.  Take  such  part  of  the  Int.  for  1  month,  as 
the  number  of  days  is  part  of  one  month. 

5.  For  years,  months,  and  days,  or  for  any  tAvo  of  these 
periods.     Find  the  Int.  for  each  period.,  and  add  the  results. 

6.  To  find  the  amount.     Add  the  Principal  to  the  Int. 

Or,  by  Proportion.  As  100  is  to  the  rate  per  cent,  so  is  the 
Principal  to  the  Int.  for  1  yr.  Then,  As  1  yr.  is  to  the  given 
time,  so  is  the  Int.  for  1  yr.  to  the  Int.  for  the  given  time. 

Note. — In  computing  Int.,  regard  30  days  1  month,  and  12 
months,  1  year.     Custom  has  made  this  lawful. 

1.  Find  the  interest  of  $360  for  2yr.,  7mon.  25 da., 
at  8  ^,  per  annum. 


operation. 
$360 
.08 


28.80  =  Int.  lyr. 
2 


6  mon.  =  | 

1  mon.  =  ,^ 

15  da.    =i 


10  da.    =i 


57. 60  =  Int.  2  yr. 
14.40  =  Int.  6  mon. 

2.40  =  Int.  1  mon. 

1.20  =  Int.  15  da. 
.80  =  Int..  lOda. 


$76.40  =  Int.  for 


operation  by  proportion. 
2yr.  7  mon.  25  da.  =  955  da. 
100    :    8    :  :    $360 


5 


360 


A  36X4     ^ 

4 =$28.80 


360  :  955  :  :  $28.80 
3?^  191 
28. gP  .40 
191 X  .40  =  $76.40  Ans. 

The    operation    may    be    further 
shortened    by   writing    the    propor- 


2yr.  7  mon.  2.5 da.  tions  together,  and  canceling. 

What  is  the  Interest  of  answers. 

2.     $350.00  for  7  yr.  3  men.,  at  4  %  ?  $101.50 


3.     $150.00  for  4yr.  2  mon.,  at  0  %  ? 
3d    Bk.  15 


$  37.50 


226  RAY'S   PRACTICAL   ARITHMETIC. 

Find  the  Interest  of  answers. 

4.  $375.40for  lyr.  8mon.,  at  6  %.  $  37.5-4 

5.  $  92.75  for  Syr.  5  mon.,  at  6  %.  $  19.01-}- 

6.  $500.00  for  lyr.  1  mon.  18  da.,  at  6%.$  34.00 

7.  $560 .  00  for  2  yr.  4  mon.  15  da.,  at  8  % .  S106 .  40 

8.  $750.00  for  4  yr.  3  mon.    6da.,at  6%. $192.00 

9.  $45G.OO  for  3yr.  5  mon.  18  da.,  at  5  %.  $  79.04 

10.  $216.00  for  5yr.  7  mon.  27  da.,  at  10%.  $122.22 

11.  $380.00  for  Syr.  9  mon.    9  da.,  at  15  %.  $215,175 

Find  the  Amount  of  answers. 

12.  $300 .  00  for  3  yr.  8  mon.,  at  6  % .  $366 .  00 

13.  $250.00  for  1  yr.  7  mon.,  at  6  %.  $273.75 

14.  $205.25  for  2yr.  8  mon.  15  da.,  at  6  %.  $238,603-}- 

15.  $150.62  for  3 yr.  5  mon.  12  da.,  at  5%.  $176,601-}- 

16.  $210 .  25  for  2  yr.  7  mon.  20  da.,  at  7  % .  $249 .  087-}- 

17.  $  57.85  for  2yr.  3  mon.  23da.,  at  5  %.  $  64.542-}- 

18.  Find  the  interest  of  $150,  from  January  9,  1847, 
to  April  19,  1849,  at  6  %.  ^;/.s.  $20.50 

Note. — To  find  the  Time  between  two  dates,  see  Art.  103. 

19.  The  interest  of  $240,  from  February  15,   1848,  to 
April  27,  1849,  at  8  %.  Aiis.  $23.04 

20.  The  interest  of  $180,  from  May  14,  1843,  to  Auirust 
28,  1845,  at  7  9^.  Ain^.  $28.84 

21.  The  interest  of  $137.50,  from  July  3,  to  Novem- 
hcr  27,  at  9  %.  Ans.  $4.95 

22.  The  amount  of  $125.40,  from  March  1,  to  Auirust 
28,  at  8.1%.  Ans.  $130.64-f 

23.  The  amount  of  $234.60,  from  Auj^ust  2,  1847,  to 
March  9,  1848,  at  5.[  %.  Aiis.  $242,024-}- 

24.  The  amount  of  $153.80,  from  Oct.  25,  1846,  to 
July  24,  1847,  at  5  %.  Am.  $159,546-}- 

Kkview. — 222.  How  find  the  interest  of  any  sum  at  any  rate  for  1  year, 
llulo  ?  For  two  or  more  years?  For  months?  For  days?  For  years, 
months,  and  days?     How  find  the  interest  hy  proiwrtion  ? 


SIMPLE   INTEREST. 


227 


ANOTHER  METHOD  FOR  INTEREST. 
Art.  223.    To  find  the  interest  of^l,at6%,  for  any  time. 

At  6  ^,  the  Int  of  $1  for  1  year,  (12  mon.)  is  6  cts.  $  .06 

For  1  mon.  it  will  be  73=^2  ^^1  ^^  ^  mills 005 

For  2  mon.,  ^X  2=1  "cent 01 

For  3  mon.,  ^X 3  =  1^  cents 015 

For  4  mon.,  ^X  4  =  2    cents.     (And  so  on.)       .    .         .02 

Again:  since  30  days  =^  1  mon. 

For  1  da.  the  Int.  is  ^^  of  ^  ct.  =  g'^  ct.  =  ^  m.   .     .     $  .000^ 

For  6  da.,  as  /o^^i'  ^*  ^^  5  ^^  5  ^*-  ^^jn  ^*-  =  1  ^^i.  .001 

Hence,  to  find  the  Interest  of  $1,  at  6  ^,  for  any  time, 

Take  as  many  cents  as  equal  half  the  even  number  of  months ; 
take  5  mills  for  each  odd  month,  1  mill  for  each  six  dar/s,  and 
one  sixth  of  a  mill  for  each  remaining  day. 

1.  Find  the  Int.  of  $1,  for  9  mon.  12  da.  at  6  %. 

Solution. —  The     preceding     shows     that     the  ^ 

Int.  for  8  months  is  4  cents;  for  1  mon.,  5  mills; 
for  12  da.,  2  mills ;  the  sum  of  these  is  $.047, 
the  required  Int.  a^^    o 

2.  Find  the  Int.  of  $1  for  17  mon.  23  da.,  at  6  % 

Solution. — For  16 mon.  the  Int.  is  Sets.;  for  1 
mon.,  5  mills ;  for  18  da.,  3  mills ;  for  5  da.,  |  of  a 
mill;  their  sura  is  %  .088g,  the  required  Int. 

3.  Find  the  Int.  of  $1  at  6  ^,  for 


.04 
.005 

.002 

.047 


.08 

.005 

.003| 

Ans.  $.088| 


16  mon. 
13  mon. 

4  mon.  18  da. 
7  mon.  12  da. 

10  mon.  13  da. 

5  mon.  17  da. 


Ans.  $ .  08 

S.065 

8.023 

$.037 

$.052 J 

$.027| 


11  mon. 

2  mon. 

9  mon. 
14  mon. 
17  mon.  27  da. 
33  mon.  20  da. 


Ans.  $. 
1  da.        $. 

3  da.        S. 

4  da.       S. 


055 
$.010| 
S.045A 
$.070i 
$.089^ 
$.168.1 


Review. — 223.  How  find  the  Int.  of  $1  at  6  per  cent,  for  any  time  ?  Why  ? 


228  RAY'S    PRACTICAL    ARITHMETIC. 

Art.  224.  To  find  the  interest  of  any  sum,  at  6  ^^^  for 
any  given  time. 

The  interest  of  $2  is  twice  as  much  as  the  Int.  of  $1  for  the 
same  time;  of  $3,  three  times  as  much,  and  so  on.     Hence,  the 

Rule  for  Finding  the  Interest  at  6  %. — Find  the  interest 
of  §1  at  6  per  cent,  for  the  given  time  (Art.  223),  then  multiphf 
this  by  the  given  sum;  the  product  will  he  the  required  Int. 

1.  Find  tlie  interest  of  869,  for  6mon.  8 da.,  at  6  ^. 

Solution.— The  interest  of  $1  for  6  mon.  8  da.,  ^  •  ^^^  1 

at  6  per  cent.  (Art.  222),  is  $.031^  ^^ 


As    the    Int.  of  $69  will   be   69  times  that  279 

of  $1,  for  the  same  time,  multiply  the  Int.  of  $1  1"" 

by  69,  to  obtain  the  result.  ^^ 


Ans.  $2.1ti2 

Find  the  Interest,  at  6  %,  of  answers. 

2.  $65.00  for    8 mon.  82.60 

3.  836.00  for  11  mon.  81.98 

4.  828.00  for    lyr.  4mon.  (16mon.)  82.24 

5.  8500.00  for    2yr.  Imon.  862.50 
G.     875.00  for    4mon.  12da.  81.65 

7.  8186.00  for    9mon.  23da.  89.083 

8.  8125.00  for    lyr.  2mon.  12da.  89.00 

9.  8210.25  for    Syr.  4mon.  24da.  842.891 

10.  8134.45  for    lyr.  5mon.  15da.  $11,764+ 

11.  8144.24  for    2yr.  3mon.  19da.  819.929+ 

Art.  225.  From  the  illustrations  in  Art.  223,  two 
rules  arc  derived:   the  2d  is  often  used. 

Rule  I. — Since  the  Int  of  $1  at  6  %,  for  Imon.  is  half  a 
cent,  to  find  the  Int.  of  any  sum  at6  ^  when  the  time  is  mon., 

Multiply  the  sum  considered  as  dollars,  by  half  the  number 
of  mon.y  the  product  icill  be  the  Int.  in  cents. 

Kkv. — 22 1.  IIow  find  tho  Int.  of  any  sum  at  6  per  cent,  for  any  time? 


SIMPLE   INTEREST.  229 

Rule  II. — Since  the  Int.  of  $1,  at  6  ^,  for  Ida.  is  ^j^  of  a 
ct.,  to  find  the  Int.  of  any  sum  at  6  %,  when  the  time  is  da., 

Multiply  the  sum  considered  as  dollars^  by  the  number  of  da., 
and  divide  the  product  by  60 ;  the  quotient  will  be  the  Int.  in  cts. 

Rem. — 1.  In  applying  Rule  1,  when  the  time  is  years  and 
months,  reduce  to  months.  When  the  time  is  days,  and  either 
months  or  years,  or  both,  find  the  Int.  for  the  days  separately, 
or,  reduce  the  whole  to  days,  and  then  find  the  Int. 

2,  Reduce  the  result  obtained  to  dollars,  by  removing  the  decimal 
point  two  places  toward  the  left.     Art.  184. 

3.  The  Int.  of  any  sum  at  5  per  cent,  is  |  of  the  Int.  at  6  pr.  ct.; 
at  7  pr.  ct.,  g  of  6  pr.  ct.,  and  so  on.  Hence,  to  find  the  Int. 
of  any  sura,  at  iany  rate,  first  find  the  Int.  at  6  per  cent.,  multiply 
this  by  the  given  rate,  and  divide  the  product  by  6.     Art.  152. 


PARTIAL   PAYMENTS   ON    BONDS,    NOTES,    &c. 

Art.  226.  When  partial  payments  have  been  made, 
the  following  rate  for  computing  the  Int.,  adopted  by 
the  Supreme  Court  of  the  U.  States,  is  regarded 

THE    LEGAL    RULE    FOR   PARTIAL   PAYMENTS. 

"  When  partial  payments  have  been  made.,  CLpply  ihe  payment, 
in  the  first  place.,  to  the  discharge  of  the  interest  then  due. 

'■^  If  the  payment  exceeds  the  Int.,  the  surplus  goes  toward 
discharging  the  principal.,  and  the  subsequent  Int.  is  to  be 
computed  on  the  balance  of  principal  remaining  due. 

"  If  the  payment  be  less  than  the  Int..,  the  surplus  of  Int. 
must  not  be  taken  to  augment  the  principal,  but  Int.  continues 
on  the  former  principal,  until  the  period  when  the  payments, 
taken  together,  exceed  the  Int.  due,  and  then  the  surplus  is 
to  be  applied  toward  discharging  the  principal;  and  Int.  is 
to  be  computed  on  the  balance,  as  aforesaid." — Kent.  C.  J. 

Rem. — A  Partial  Payment,  is  payment  of  part  of  a  note,  or 
other  obligation,  which  payment  should  be  indorsed  upon  it. 

The  rule  is  founded  on  the  principle,  that  neither  interest  nor 
payment  shall  draw  interest. 


230  RAY'S   PRACTICAL   ARITHMETIC. 

1.  $350.  Boston,  July  1st,  1845. 

For  value  received,  I  promise  to  pay  to  Edward  Sargent, 
or  order,  on  demand.  Three  hundred  and  fifty  dollars, 
with  interest  at  6  %.  Lowell  Mason. 

On  this  note  the  following  payments  -were  indorsed : 

March  1st,  184G,  rec'd  $44.  Jan.  1st,  1847,  rec'd  $26. 

Oct.       1st,  1846,  rec'd  §10.  Dec.  1st,  1847,  rec'd  $15. 

What  was  due,  on  settlement,  March  IGth,  1848? 

OPERATION. 

Principal, $350.00 

Interest  to  1st  payment  (8  mon.), 14.00 

Amount  due  March  1st,  1846, 364.00 

1st  payment  (greater  than  Int.)  deducted,  ....  44.00 

Balance  due  March  1st,  1846, $320.00 

Int.  on  bal.  to  Oct.  1st,  1846,  (7  mon.)     .     .     $11 .20 
2d  payment,  (less  than  Int.  due,)    ....        10.00 

Surplus  Int.  unpaid,  Oct.  1st,  1846,     ...  1.20 

Int.  continued  on  bal.  from  Oct.  1st,  1846,  to  ) 

Jan.  1st,  1847,  (3  mon.) |     4.80        fi.OO 

Amount  due  Jan.  1st,  1847, 320.00 

Deduct  3d  payment,  (greater  than  Int.  due,)  ...  26.00 

Balance  due  Jan.  1st,  1847, $300.00 

Int.  on  bal.  to  Dec.  1st,  1847,  (11  mon.)  .     .     $16.50 
4th  payment,  (less  than  Int.  due,)  ....        15.00 


Surplus  Int.  unpaid  Dec.  Isf,  1847,     ...  1.50 

Int.  continued  on  bal.  from  Dec.  1st,  1847,  to  ) 

March  lOih,  1848,  (3  mon.  15 da.)    .     .     .    )      5.25        6.75 

Balance  due,  on  settlement,  March  16th,  1848,     .     .     $306.75 

SuGGE.sTiON, — When  any  payment  is  less  than  the  Interest  on 
the  principal  or  balance  up  to  the  time  of  payment,  shorten  the 
operation  by  deferring  the  computation  of  Int.  until  the  payments, 
taken  together,  exceed  tlie  Int.  due.  Thus,  in  the  above  operation, 
instead  of  computing  the  Int.  Oct.  1st,  1846,  it  might  have  been 

Review. — 225.  When  the  rate  is  6  per  cent.,  how  find  tho  interest 
when  tho  time  is  months  ?  When  days  ?  Kem.  How  may  tho  interest 
at  any  rate  bo  found  by  Rules  1  and  2?     Why  ? 

226.  What  is  tho  legal  rule  for  partial  payments  ? 


SIMPLE    INTEREST.  231 

deferred  till  Jan.  1st,  1847.    The  whole  Int.  then,  would  have  beea 
$10,  which,  added  to  §320,  the  preceding  balance,  =$330. 

2.  A  note  of  $200  is  dated  Jan.  1,  1845,  on  which 
there  is  paid,  Jan.  1,  18-16,  $70  :  reckoning  interest  at 
G  %,  what  was  due  Jan.  1,  1847?  Ans.  $150.52 

3.  A  note  of  $300  is  dated  July  1st,  1843.     Int.  6  %, 
Indorsed,  Jan.  1,  1844,  $109.  July  1,  1844,  $100. 
What  was  due  Jan,  1st,  1845?  ^»s.$109.18 

4.  A  note  of  $150  is  dated  May  10th,  1850.     Int.  6  %, 
Indorsed,  Sept.  10,  1851,  $32.  Sept.  10,  1852,  $6.80 
What  was  due  Nov.  10th,  1852?      Ans.  $132.30 

5.  A  note  of  $200  is  dated  Mar.  5th,  1841.     Int.  10  %. 
Indorsed,  June  5,  1842,  $20.  Dec.  5,  1842,  $50.50 
What  was  due  June  5th,  1844?       Ans.  $189,175 

6.  A  note  of  $250  is  dated  Jan.  1st,  1845.     Int.  7  %. 
Indorsed,  June  1,  1845,  $6.  Jan.  1,  1846,  $21.50 
AVhat  was  due  July  1st,  1846  ?  Ans.  $248.40 

7.  A  note  of  $180  is  dated  Aug.  1st,  1844.     Int.  6  %. 

Indorsed,  Feb.  1,  1845,  $25.40  Aug.  1,  1845,  $4.30 

Jan.  1,  1846,  $30. 

What  was  due  July  1st,  1846  ?        Ans.  $138,535 

8.  A  note  of  $400  is  dated  March  1st,  1845.    Int.  6  %, 

Indorsed,  Sept.  1,  1845,  $10.  Jan.  1,  1846,  $30. 

July  1,  1846,  $11.  Sept.l,  1846,  $80. 

What  was  due  March  1st,  1847?      Ans.  $313,326 

9.  A  note  of  $450  is  dated  April  16th,  1846.     Int.  8  %. 

Indorsed,  Jan.    1,  1847,  $20.  April  1,  1847,  $14. 

July  16,  1847,  $31.  Dec.  25,  1847,  $10. 

July    4,  1848,  $18. 

What  was  due  June  1st,  1849?  Ans.  $466.50 

10.  A  note  of  $1000  is  dated  Jan.  1st,  1840.     Int.  6  %. 

Indorsed,  May     1,  1840,  $  18.  Sept.     4,  1840,  $  20. 

Dec.  16,  1840,  $  15.  April  10,  1841,  $  21. 

July  13,  1841,  $118.  Dec.    23,  1841,  $324. 

What  was  due,  Oct.  1st,  1843?  Ans.  $663.80 


232  RAY'S   PRACTICAL    ARITHMETIC. 

Art.  227.  In  Computing  Interest,  when  partial 
PAYMENTS  have  been  made,  the  following  rule  is  applied 
where  settlement  is  made  in  a  year,  or  in  less  than  a  ycar^ 
from  the  commencement  of  Int. 

Find  the  amount  of  the  principal  for  the  whole  time ;  then 
find  the  amount  of  each  payment  from  the  time  it  was  made  to 
the  time  of  settlement.  Add  together  the  amounts  of  the  several 
payments^  and  subtract  their  sum  from  the  amount  of  the  prin- 
cipal; the  remainder  will  he  the  sum  due  on  settlement. 

1.  A  note  of  $320  is  dated  Jan.  1st,  1846.     Int.  6  %. 

Indorsed,  May  1,  1846,  $50.  Nov.  16,  1846,  $100. 

What  was  due  Jan.  1st,  1847?  Ans.  $186.45 

2.  A  note  of  $540  is  dated  March  1st,  1847.     Int.  8  %. 

Indorsed,  May  1,  1847,    $90.  July  1,  1847,  $100. 

Aug.  1,  1847,  $150.  Oct.  11,  1847,  $180. 

What  was  due  Jan.  1st,  1848?  Am.  $39. 

Art.  223.  Connecticut  rule. 

"Compule  the  interest  to  the  time  of  the  first  payment;  if  that 
be  1  year  or  more  from  the  time  the  Int.  commenced,  add  it  to  the 
principal,  and  deduct  the  payment  from  the  sum  total. 

"  If  there  be  after  payments  made,  compute  the  Int.  on  the  bal- 
ance due  to  the  next  payment,  and  deduct  the  payment  as  above; 
and,  in  like  manner,  from  one  payment  to  another,  till  all  the  pay- 
ments are  absorbed,  provided  tlie  time  between  one  payment  and 
another  be  1  year  or  more.     But, 

"If  any  payments  be  made  before  1  yr.'s  Int.  has  accrued,  then 
compute  the  Int.  on  the  principal  sum  due  on  the  obligation,  for  1  yr., 
add  it  to  the  Prin.,  and  compute  the  Int.  on  the  sum  paid,  from  the 
time  it  was  paid  to  the  end  of  the  yr.,  add  it  to  the  sum  paid,  and 
deduct  that  sum  from  the  Prin.  and  Int.  added  as  above.* 

"  If  any  payments  be  made  of  a  less  sura  than  the  Int.  arisen  at 


*If  a  year  docs  not  extend  beyond  the  time  of  payment;  but  if  it 
does,  then  find  the  amount  of  the  principal,  remaining  unpaid,  up  to  the 
time  of  settlement,  likewise  the  amount  of  the  payment  or  payments  from 
the  time  they  were  paid  to  the  time  of  settlement,  and  deduct  the  sum  uf 
those  several  amounts  from  the  amount  of  the  principal. 


SIMPLE    INTEREST.  233 

the  time  of  such  payment,  no  Int.  is  to  be  computed,  but  only  on 
the  principal  sum  for  any  period." — Kirby's  Keports. 

1.  A  note  of  3875  is  dated  Jan.  10th,  1831.     Int.  6  %. 
Indorsed,  Aug.  10,  1834,  $260.  Dec.  16,  1835,  $300. 

Mar.    1,  1836,  $  50.  July    1,  1837,  $150. 

What  was  due  Sept.  1st,  1838?      Ans.  ^46.983-(- 

PROBLEMS    IN   INTEREST. 

Art.  229.  There  are  fo^ir  parts  or  quantities  connected 
with  each  operation  in  Interest :  these  are,  the 

Principal:  Rate  per  cent. :   Time:  Interest  or  Amount. 

If  any  three  of  them  are  given,  the  other  may  be  found. 

Problem  I. — To  find  the  interest  or  amount,  the 
Principal,  Rate  per  cent,  and  Time  being  given. 

This,  the  most  important  problem  in  Interest,  is  illustrated 
in  Art.  220  to  223. 

Art.  230.  Problem  II. — To  find  the  time,  the  Prin- 
cipal, Rate  per  cent,  and  Interest  being  given. 

1.  I  loaned  3200  at  6  ^,  and  received  336  for  interest: 
how  long  was  the  money  loaned  ? 

Solution.— The  interest  of  $200  (Art.  220),  operation. 

at  6  per  cent.,  for  1  year,  is  $12 ;  then,  since  the  3200 

given  principal  in  1  year  produces  $12  Int.,  it  •  06 

will  require  the  same  principal  as  many  times  1  

year  to  produce  $36  Int.,  as  $12  are  contained  1-^.00 

times  in  $36;  that  is,  8,     Ans.  3  yr.  336 -=-12=:  3. 

*2.  In  what  time  will  360,  at  5  %,  gain  312  interest? 

Ans.  4yr. 

RULE   FOR   PROBLEM   II. 

Divide  the  given  interest  hy  the  Int.  of  the  principal  at  the 
given  rate  per  cent,  for  one  year ;  the  quotient  will  be  the  time. 

Or,  By  Proportion. — As  the  Int.  of  the  principal  for  one 
year  is  to  the  given  Int.,  so  is  one  year  to  the  required  time. 

Review. — 229.  What  quantities  are  connected  with  each  operation  in 
interest  ?  How  many  must  be  given,  that  the  remaining  may  be  found  ? 
What  is  the  most  important  problem  in  interest  ? 

230.  What  is  Problem  II  ?    What  the  rule  ?    How,  by  proportion  ? 


234  RAY'S    PRACTICAL    ARITHMETIC. 

Note. — When  the  quotient  contains  a  common  fraction,  or  a 
decimal,  find  its  value  in  months  and  days.     Art.  162  and  188. 

3.  A  man  loaned  S375,  at  8^,  and  received  $00 
interest:  how  long  was  it  loaned?  Ans.  3yr. 

4.  How  long  will  it  take  $225,  at  4  5^,  to  gain  $66 
interest?  Ans.  7 1  yr.  =  7yr.  4mon. 

5.  How  long  will  it  take  8250,  at  6%,  to  gain 
S34.50  interest?  Am.  2.3yr.  =  2yr.  3mon.  18da. 

6.  How  long  will  it  take  $60,  at  6%,  to  gain  $13.77 
interest?  Ans.  3.825yr.  =  3yr.  9m"on.  27 da. 

7.  $500,  at  10  %,  to  amount  to  $800?         Ans.  6yr. 

Suggestion. — By  subtracting  the  principal,  §500,  from  the 
amount,  $800,  the  remainder,  $300,  is  the  interest  of  the  principal 
for  the  required  time :   then  proceed  as  before. 

8.  In  what  time  will  $G00,  at  9%,  amount  to  $798? 

Ans.  3yr.  8mon. 

9.  In  what  time  will  $200.  at  6%,  amount  to  $400; 
or,  in  what  time  will  any  sum  of  money  double  itself, 
at  6^?  ^«s.  16  yr.  8mon. 

SuGGESTTO.v. — To  find  in  what  time  any  principal  will  double 
itself  at  any  given  rate  per  cent, 

Take  any  sitm,  as  $100,  /or  a  principal,  and  find  the  time  in  which, 
nt  the  givenrate  per  cent.,  the  Int.  produced  would  equal  the  principal. 

IN  WHAT  TIME  WILL  ANY  PRINCIPAL  DOUBLE  ITSELF, 

10.  At  4  ^  per  annum?  Ans.  25 yr. 

11.  At  5  %  ?  Ans.  20   yr. !  14.  At  10  <^  ?  Ans.  lOyr. 

12.  At  7  %  ?  Ans.  14^  yr.    15.  At  12  %  ?  Ans.  ^  yr. 

13.  At  8  %?  Ans.  12.5yr.   16.  At  16%?  Ans.  6]yr. 

17.  In  what  time,  at  15  %,  will  any  principal  treble 
itself?      Ans.  \^  yr.       Quadruple  itself?       Ans.  20 yr. 

Art.  231.  Problem  III.  To  find  the  rate  per 
CENT.,  the  Principal,  Interest,  and  Time  being  given. 

Review. — 230.  IIow  find  tho  timo  in  which  any  principal  will  double 
itself,  at  any  given  rate  per  cent.  ? 

231.  What  is  problem  III  ?     What  the  Rule  ?    How,  by  proportion  ? 


SIMPLE    INTEREST.  235 

1.  I  borrowed  8600  for  2  years,  and  paid  848  interest : 
what  rate  ^^  did  I  pay? 

OPERATION. 

Solution.— The  Int.   of  $600  for  1  $600X.01  =  $6. 

year,  at  1  per  cent.  (Art.  220),  is  $6, 
and  for  2  years,  $6X2  =  $12.  $6X2  =  $12  =  Int    for 

As  the  given  Int.  ($48),  contains  this  ^^^-  ^<^  -^  ^• 

Int.  ($12),    exactly   4  times,   the  rate  $48  -j-  12  =  4  ^  Ans. 

must  have  been  4  times  as  large;  that  is  4  per  cent. 

*2.  A  man  paid  828  for  the  use  of  880  for  5  years : 
what  was  the  rate  %  ?  Ans.  7. 

RULE  FOR  PROBLEM  III. 

Divide  the  given  interest  by  the  Int.  of  the  principal  atl  per 
cent,  for  the  given  time ;  the  quotient  will  he  the  rate  per  cent. 

Or,  by  Proportion.  As  the  Int.  of  the  principal  at  1  per 
cent,  for  the  given  time,  is  to  the  given  Int.,  so  is  1  per  cent 
to  the  rate  per  cent. 

3.  A  merchant  paid  830  for  the  use  of  8300  for  lyr. 
8mon. :  find  the  rate  ^.  Ans.  G. 

4.  A  broker  paid  8200  for  the  use  of  81000  for  2yr. 
6mon,  :  find  the  rate  %.  Ans.  8. 

5.  823.40  was  paid  for  the  use  of  8260  for  2  years: 
what  the  rate  %  ?  Ans.  4i. 

6.  8110.40  was  paid  for  the  use  of  8640  for  6  years : 
find  the  rate  % .  Ans.  2 .  875  =  2|. 

Art.  232.  Problem  TV.  To  find  the  principal,  the 
Time,  Kate  per  cent.,  and  Int.  being  given. 

1.  What  sum  in  2yr.,  at  6  %,  will  gain  827  Int.? 

Solution. — The  interest  of  $1  for  2  years,  at  6  per  cent., 
is  12  cents,=8.12  Since  12  cts.  is  the  Int.  of  $1  for  the  given 
time  and  rate,  $27  must  be  the  Int.  of  as  many  times  $1,  as  12  cts. 
are  contained  times  in  $27.     $27 -^  $-12  =  225.     Ans.  $225. 

*2.  What  sum  of  money  in  3yr.,  at  5  %,  will  it  take 
to  j'ain  88.25  Int.?  Ans.  855. 


23G  RAY'S   PRACTICAL   ARITHMETIC. 

RULE    FOR   PROBLEM    IV. 

Divide  the  given  interest  hy  the  Int.  of  $1  for  the  given  time 
at  the  given  rate  per  cent. ;  the  quotient  will  be  the  principal. 

Or,  by  Proportion.  As  the  Lit.  of  $1  for  the  given  time,  at 
the  given  rate  per  cent,  is  to  the  given  Int.,  so  is  $1  to  the  re- 
quired principal. 

What  Principal, 

3.  At  5  %,  will  gain  $341.25  in  3  jr.?    Ans.  $2275. 

4.  At  6  %,  will  gain  $2.26  in  16  men.?    Ans.  $28.25 

5.  At  5  ^,  will  produce  a  yearly  interest  of  $1023.75? 

Ans.  $20475. 

6.  At  9  %,  will  gain  $525,398  in  12  yr.  3  mon.  20  da.? 

Ans.  $474.40 

Problem  V.  To  find  the  principal  and  interest, 
the  Time,  Rate  per  cent.,  and  Amount  being  given. 

This  is  treated  of  under  Discount.     See  Art.  238. 

Art.  233.  compound   interest 

Is  interest  on  the  principal,  and  also  on  the  interest 
itself,  after  the  latter  becomes  due. 

(Simple  Interest  is  interest  on  the  principal  only.) 

1.  Find  the  compound  Int.  of  $300  for  3  yr.  at  6  %. 

$300.   Principal  1st  year. 
$300X.0G=      18.   Interest     1st  year. 

$318.        Principal  2d  year. 
$318 X  .06=      19.08  Interest     2d  year. 

$337 .  08       Principal  3d  year. 
$337 .  08  X  .  06=      20 .  2248  Interest     3d  year. 

$357.3048  Amount  for  3  yr. 
300.  Given  principal. 

$  57.3048  Compound  Int.  for  3  yr. 

*2.  Find  the  compound  Interest  of  $200  for  2  years, 
at  8%.  ^»ts.  $33.28 


Review.— 2.32.  What  is  Pn.blcm  IV  ?    What  tho  Rulo  for  Prob.  IV  ? 


COMPOUND   INTEREST.  237 

RULE    FOR   COMPOUND    INTEREST. 

Find  the  amount  of  the  given  principal  (Art.  222)^*0?'  the  \st 
yr.,  and  make  it  the  principal  for  the  'Id  yr. 

Find  the  amount  of  this  principal  for  the  2d  yr.,  make  it  the 
principal  for  the  3d  yr.,  and  so  on,  for  the  given  number  ofyr. 

From  the  last  amount  subtract  the  given  principal ;  the  re- 
mainder will  be  the  compound  interest. 

Notes. — 1.  When  the  interest  is  payable  half-yearly,  or  quarter- 
yearly,  find  the  Int.  for  a  half,  or  a  quarter  year,  and  proceed  in 
other  respects  as  when  the  Int.  is  payable  yearly. 

2.  When  the  time  is  years,  months,  and  days,  find  the  amount  for 
the  years,  then  compute  the  Int.  on  this  for  the  months  and  days, 
and  add  it  to  the  last  amount. 

Find  the  Amount,  at  6  ^,  Compound  Interest, 

3.  Of  $500  for  3  years Am.  $595,508 

4.  $800  for  4  years Am.  $1009. 98-|- 

Find  the  Compound  Interest 

5.  Of  $250  for  3  yr.,  at  6  %.  .     .     .     Ans.  $47,754 

6.  $300  for  4yr.,  at  5  %.  .     .     .     JL^is.  $64.65 -f- 

7.  $200  for  2  yr.,  at  6  %,  payable  semi-annually. 

Ans.  $25,101  + 

8.  Find  the  amount  of  $500  for  2  yrs.,  at  20  %  com- 
pound interest,  payable  quarterly.  Ans.  $738,727  + 

9.  What  is  the  compound  interest  of  $300  for  2  yr.  6 
mon.,  at  6  %  ?  Am.  $47,192+ 

10.  What   the   compound  interest  of  $1000  for  2  yr.  8 
mon.  15  da.,  at  6  %  ?  Am.  $171,353 

Art.  234.  Since  the  amount  of  $2,  for  any  given  time, 
will  be  twice  the  amount  of  $1 ;  the  amount  of  $3,  three 
times  as  much,  &c.;  Therefore, 

If  a  table  be  formed  containing  the  amounts  of  $1  for  1,  2,  3, 
&c.,  years,  any  of  these  amounts  multiplied  by  a  given  Prin- 
cipal will  give  its  amount  at  Compound  Interest  for  the  same 
time  and  rate.     See  Table,  page  238. 


238 


RAYS   PRACTICAL   ARITHMETIC. 


TABLE 

Showing  the  amount  of  $1,  at  3,  4,  5,  6,  7  and  8  per  cent.,  Com- 
pound Interest,  for  any  number  of  years,  from  1  to  25. 


Yr. 

3  per  cent. 

4  per  cent. 

5  per  cent. 

G  per  cent. 

7  per  cent. 

8  per  cent. 

1 

1.03 

1.04 

1.05 

1.06 

1.07 

1.08 

2 

1.060 

1.0816 

1.1025 

1.1236 

1.1449 

1.1664 

3 

1.092727 

1.124864 

1.157625 

1.191016 

1.225043 

1.259712 

4 

1.12551)9 

1.169859 

1.215506 

1.262477 

1.310796 

] .360488 

5 

1.159274 

1.216653 

1.276282 

1.338226 

1.402551 

1.46932S 

6 

1.194052 

1.265319 

1.340096 

1.418519 

1.500730 

1.586874 

7 

1.229874 

1.315932 

1.407100 

1.503630 

1.605781 

1.713824 

8 

1.26G770 

1.368569 

1.477455 

1.593848 

1.718186 

1.850930 

9 

1.304773 

1.423312 

1.551328 

1.689479 

1.838459 

1.999004 

10 

1.343916 

1.480244 

1.628895 

1.790848 

1.967151 

2.158924 

11 

1.384234 

1.539454 

1.710339 

1.898299 

2.104851 

2.331638 

12 

1.425761 

1.601032 

1.795856 

2.012196 

2.252191 

2.518170 

13 

1.468534 

1.665074 

1.885649 

2.132928 

2.409845 

2.719623 

14 

1.512590 

1.731676 

1.979932 

2.260904 

2.578534 

2.937193 

15 

1.557967 

1.800944 

2.078928 

2.396558 

2.759031 

3.172169 

16 

1.604706 

1.872981 

2.182875 

2.540352 

2.952163 

3.425942 

17 

1.652848 

1.947900 

2.292018 

2.692773 

3.158815 

3.700018 

18 

1.702433 

2.025817 

2.406619 

2.854339 

3.379932 

3.996019 

19 

1.753506 

2.106849 

2.526950 

3.025600 

3.616527 

4.315701 

20 

1.806111 

2.191123 

2.653298 

3.207135 

3.869684 

4.660957 

21 

1.860295 

2.278768 

2.785963 

3.399564 

4.140562 

5.033833 

22 

1.916103 

2.369919 

2.925261 

3.603537 

4.430401 

5.436540 

23 

1.973587 

2.464716 

3.071524 

3.819750 

4.740529 

5.871463 

24 

2.032794 

2.563304 

3.225100 

4.048935 

5.072366 

6.341180 

25 

2.093778 

2.665836 

3.386355 

4.291871 

5.427432 

6.848475 

11.  What,  by  the  Table,  will  be  the  Amount  of  370  for 
9yr.,  at  5  ^  compound  interest? 

Solution. — By  the  Table,  the  amount  of  SI  for  9yrs.,  at  5  pr.  ct., 
is  $1.551328;  and  $1.551328X70  =  $108.5929G 

12.  Of  S345  for  10  yr.,  at  G  %  ?         Ans.  $617.81-h 

13.  S200  for  41  yr.,  at  G  %  ? 

ExPL.\NATiON. — Take  any  two  periods  whose  sum  is  41  years, 
thus,  20yr.-|-21  yr.=41  yr.,  and  find  the  amount  for  the  1st  period: 
then  regard  this  a  new  principal,  and  find  its  amount  for  the  2d 
period  :   (lie  last  amount  will  be  the  Ans. 

The  Tabular  number  for  20  yr.  is  3.207135;  this,  multiplied 
by  200,  gives  $641,427,  amount  for  20  yr.:  Tabular  number  for  21  yr. 
is  3.300504,  which X041.427  =  $2180.572-1-  Atis. 


DISCOUNT.  23a 

What,  by  the  Table,  will  be  the  Interest 

14.  Of  $890  for  30  yr.,  at  6  %  ?  Ans.  $4221.70-}- 

15.  $200  for  70  yr.,  at  5  %  ?  Ans.  $5885.28+ 

Art.  235.   DISCOUNT 

Is  a  deduction  made  for  the  payment  of  money  before  it 
is  due.     For  example, 

If  a  debt  of  $106,  due  one  year  hence  without  interest, 
be  paid  at  the  present  time,  the  sum  paid,  with  one  years 
interest  added,  should  make  $106.     And, 

If  the  rate  per  cent,  is  6,  this  sum  would  be  $100  ; 
for,  the  amount  of  $100  at  interest  for  1  year,  at  Q  %, 
is  $106.     Art.  220. 

The  PRESENT  WORTH  of  a  debt  payable  at  a  future 
time  without  interest,  is  that  sum  which,  at  a  specified 
rate  ^  for  the  same  time,  would  amount  to  the  debt. 

The  DISCOUNT  is  the  sum  deducted  for  present  payment. 

Art.  236.  1.  Find  the  Present  Worth  of  $224,  due 
2  yr.  hence,  without  interest,  money  being  worth  6  ^ 
per  annum. 

SoLUTiox. — The  amount  of  $1  for  2  years,  at  6  per  cent.,  is 
$1.12;  henc€,  the  present  worth  of  each  $1.12  of  the  given  sum, 
is  $1.  And,  the  present  worth  of  $224,  will  be  as  many  times 
$1,  as  $1.12  is  contained  times  in  $224. 

$224.-^$1.12  =  200.     Ans,  $200. 

Proof. — The  amount  of  $200  for  2  years,  at  6  per  cent., 
is  $224.     Art.  220. 

*2.  Find  the  present  worth  of  $81,  due  2yr.  hence, 
no  Int.,  money  worth  4  ^  per  annum.  Ans.  $75. 

Eeview.— 234.  What  does  the  Table  show  ?  By  means  of  it,  how  find 
the  amount  of  any  sum  at  compound  interest  ? 

285.  What  is  discount?  What  the  present  worth  of  a  debt,  payable  at 
a  future  timo  without  interest  ?     The  discount? 


240 


KAY'S    PRACTICAL    ARITHMETIC. 


OP    PRESENT    WORTH. 

To  find  the  present  worth  of  a  sum  payable  at  a  future 
time  without  interest  : 

Rule. — Divide  tlie  given  debt  by  the  amount  of  one  dollar  for 
the  given  time,  at  the  given  rate  per  cent. ;  the  quotient  will  be 

the    PRESENT   -WORTH. 

To  find  the   discount,  subtract  the   present  worth  from 

the  DEBT. 

Or,  by  Proportion.  As  the  amount  of  any  jyrincipal,  (as$l, 
or  $100,)  for  the  given  time.,  at  the  given  rate  per  cent,  is  to  the 
principal,  so  is  the  given  debt  to  the  present  worth. 

Note. — In  the  following  Examples,  the  rate  per  cent,  is  6,  un- 
less some  other  is  given ;  and  when  the  present  worth  of  any  sum 
is  required,  it  is  supposed  not  to  be  at  interest. 


What  is  the  Present  Worth  of 

3.  $1300.00  due  5  years  hence?     . 

4.  $4720.00  due  3  years  hence?     . 

5.  $257.50  due  1  year    hence?     . 

6.  $199.80  due  1  yr.  lOmon.  hence 

7.  $675.00  due  5  yr.  10  mon.  hence 

8.  $307.50  due  5  mon.  hence?     . 

9.  $493.20  due  in  7yr.  9 mon.  20 da 

What  is  the  Discount  of 

10.  $496.00  due  4  years  hence?  .     . 

11.  $276.64  due  2  years  hence?  .     . 


ANSWERS. 

.  $1000. 
.  S4000. 
$242.9244- 

$180. 

$500. 

$300. 
$335.89+ 


$96.00 
$29.64 
$55.00 


12.  $330.00  due  3yr.  4 mon.  hence?     . 

13.  Bought   $260  wortli  of  jroods,   on  8  mon.   credit: 
what  sum  will  pay  the  debt  now?  Ans.  $250. 

14.  If   money   is   worth    12  ^,    what    is    the   present 
worth  of  $235.20,  due  1  year  hence?  Ans.  $210. 

15.  What  is  the  discount,  at  7%,  of  $401.25,  due  1 
year  hence?  Am.  $26.25 


DISCOUNT.  241 

16.  "What  is  the  difference  between  the  simple  Int.  and 
discount  of  $1080  for  10  yr.,  at  6  %  ?  Ans.  ^243. 

17.  A  man  was  offered  $1122  for  a  house,  in  cash,  or 
$1221,  payable  in  lOmon.  without  Int.  He  chose  the 
latter :  how  much  did  he  lose,  supposing  the  note  dis- 
counted at  the  rate  of  12  ^  per  annum  ?  Ans.  $12. 

Art.  237.     payments  at  different  times. 

When  payments  without  interest,  are  to  be  made  at 
different  times,  to  find  the  present  value  of  the  whole, 

Find  the  present  worth  of  each  payment,  and  take  their  sum. 

18.  Find  the  present  value  of  a  debt  of  $956.34,  one- 
third  to  be  paid  in  lyr.,  one-third  in  2}^.,  and  one-third 
in  3yr.;  money  being  worth  5  ^.  Ans.  $870.60 

19.  Of  a  debt  of  $1440,  of  which  one-half  is  payable 
in  3mon.,  one-third  in  6mon.,  and  the'  remainder  in  9 
mon. ;  Int.  at  6  %  per  annum.  Ans.  $1405.044-}- 

20.  Of  a  debt  of  $700,  of  which  $60  are  to  be  paid  in 
6 mon.,  S180  in  lyr.,  $260  in  18 mon.,  the  remainder 
in  2yr. ;  Int.  6  %  per  annum.  Ans.  $645,167-1- 

DISCOUNT   AND   INTEREST    COMPARED. 

Art.  238.  A  comparison  of  Discount  with  Interest 
(Art.  219),  shows  that  the  Present  Worth  corresponds  to 
the  principal,  the  debt  to  the  amount,  and  the  discount  to 
the  interest  of  the  principal  for  the  given  time,  at  the 
given  rate  per  cent. ;  hence. 

When  the  time,  rate  per  cent.,  and  amount  are  given,  the 
principal  is  found  (Art.  230)  by  dividing  the  amount  by  the 
amount  q/"  $1  for  the  given  time,  at  the  given  rate  per  cent. 

To  fnd  the  interest,  subtract  the  principal  from  the  amount. 

1.  The  amount  is  $650;  time,  5yr. ;  rate,  6  ^  :  what 
is  the  principal?  Ans.  $500. 

Eeview. — 236.  How  find  the  present  worth,  Eule ?  The  discount? 
How,  by  proportion  ?  237.  When  payments  without  interest  are  to  bo 
made  at  different  times,  how  find  the  present  value  of  the  whole  ? 

3d  Bk.  10 


242  RAY'S  PRACTICAL  ARITHMETIC. 

2.  What  principal,  at  interest  for  9  yr.,  at  5%,  will 
amount  to  $725  ?  Ans.  $500 . 

3.  The   amount  is   $571.20;  time,  4yr.;   rate,  5  %: 
what  is  the  interest?  Ans.  $95.20 

4.  A  note,  at  interest  for  2yr.  6mon.,  at  G  ^,  amounts 
to  $690  :  find  the  interest.  Ans.  $90. 


Art.  239.  BANK  DiscGUiiT. 

A  PROMISSORY  NOTE  is  a  written  promise  by  one  or 
more  persons  to  pay  to  another,  a  named  sum  of  money, 
after  a  specified  time  has  elapsed  from  its  date. 

A  note  is  discounted  when  a  bank  receives  it,  and 
pays  to  the  holder  what  remains  after  deducting  from 
its  face  the  interest  on  it,  till  it  becomes  due. 

Bank  discount  is  the  interest  deducted  from  the  face  of 
the  note. 

The  hank  discount^  therefore,  is  the  Simple  Interest  of  tJie 
FACE  of  the  note  paid  in  advance. 

The  time  to  elapse  from  any  given  date  till  a  note  becomes 
due,  is  termed  days  to  run.  By  usage,  a  note  or  draft  is  not 
really  due  till  th-ee  days  after  the  time  specified  for  payment. 

These  three  days  are  called  days  of  grace,  but  banks  charge 
interest  for  them. 

Hence,  in  calculating  the  interest,  three  days  must  be  added 
to  the  time  specified  in  the  note. 

The  FACE  of  a  note  is  the  sum  promised  to  be  paid;  the  tko- 
CEEDS,  the  sum  realized  when  the  note  is  discounted. 

Rem. — 1.  Before  a  bank  discounts  a  note,  it  is  required  that  one 
or  more  persons  shall  indorse  it ;  that  is,  -write  their  names  upon 
the  back,  by  which  they  become  responsible  for  its  payment. 

Eeview. — 238.  Comparing  discount  with  interest,  to  what  docs  tho 
present  worth  correspond  ?  Tho  debt  ?  Discount  ?  When  tho  time,  rato 
per  cent.,  and  amount  are  given,  how  find  tho  principal  ?    Tho  interest? 

239.  What  is  a  promissory  note  ?  What  is  bank  discount  ?  What  n'-o 
dtiya  to  rim?  When  is  a  noto  rcaUi/  duo?  What  aro  days  of  grace ? 
What  is  the  face  of  a  noto  ?    What  tho  proceeds  ? 


BANK   DISCOUNT.  243 

2.  Bank  discount  is  diflFerent  from  true  discount.  Art.  235. 
The  bank  discount  of  $106  for  1  year,  is  $6.36,  while  the  true 
discount,  Art.  236,  is  $6.  The  diflFerence,  36  cents,  is  the  interest 
of  the  true  discount  for  the  same  time. 

For  a  fuller  discussion  of  this  subject,  see  "  Ray's  Hlglier  Aritlim£tic" 

Art.  240.  In  bank  disct)unt,  as  interest  is  always  to 
be  computed  for  three  days  more  than  the  specified  time 
of  payment,  the  calculations  involve  the  finding  of  Int. 
for  days ;  see  Rule  2,  Art.  225. 

The  rate  per  cent,  is  always  6,  unless  some  other  is  given. 

1.  What  is  the  bank  discount  and  proceeds  of  a  note 
of  $100,  payable  60  days  after  date  ? 

OPERATION. 

Face  of  the  note $100.00 

Int.  of  $100  for  63  da.,  (Art.  225)       1 .  05  bank  discount. 

An%.  $98.95  proceeds. 

Find  the  Bank  discount  of  a  note  of  answers. 

2.  $137,  payable  90  days  after  date.  $2.12-1- 

3.  $1780,  payable  90  days  after  date.  $27.59 

4.  $375,  payable  30  days  after  date.  $2.06| 

5.  $165,  payable  60  days  after  date.  $1,731 

6.  $140,  payable   4  mon.  after  date.  $2.87 

7.  $80,  payable    6  mon.  after  date.  $2.44 

Find  the  Proceeds  or  Avails  of  a  note  of 

8.  $180,  due  30  days  after  date.  $179.01 

9.  $960,  due  30  days  after  date.  $954.72 

10.  $875,  due  90  days  after  date.  $861. 43| 

11.  $3900,  due  60  days  after  date.  $3859.05 

Review. — 239,  Eem.  1.  "What  is  meant  by  indorsing  a  note?  2.  Are 
bank  discount  and  true  discount  the  same  ?  Show  their  difiFerence  by  an 
example.    240.  By  what  rule  is  bank  discount  generally  calculated  ? 


244  RAY'S    PRACTICAL    ARITHMETIC. 

12.  Find  the  proceeds  of  a  note  of  $2580,  due  100  days 
after  date,  discount  b%.  Ans.  S2543.09-|- 

13.  I  bought  225  barrels  flour,  at  83.50  per  bl. ;  sold 
it  at  $4:  per  bl.,  taking  a  note  payable  6  mon.  after  date. 
If  this  note  be  discounted  at  G  %  per  annum,  what  the 
gain  by  the  transaction  ?  Ans.  S85.05 

14.  What  the  difference  between  true  discount,  and 
bank  discount,  of  $535,  for  1  year,  at  7  ^,  not  reckoning 
days  of  grace  ?  Ans.  $2 .  45 

15.  Omitting  the  3  days  of  grace,  find  the  difference 
between  the  true  and  bank  discount  of  $1209,  for  4  yr., 
at  6  ^  per  annum.  Ans.  $56.16 

Art.  241.  To  make  a  note,  the  proceeds  of  ivhich,  when 
discounted,  shall  be  a  given  sum. 

1.  For  what  sum,  due  90  days  hence,  must  I  give  a 
note,  that  when  discounted  at  6  ^  per  annum,  the  pro- 
ceeds will  be  $177.21  ? 

Sol.— The  bank  discount  of  $1  for  93  d<a.,  is  $.0155  (Art.  223); 
which,  deducted,  from  $1,  leaves  $.9845,  the  proceeds  of  a  note 
of  $1,  discounted  for  the  same  time.  Therefore,  for  each  $.9845  of 
proceeds,  the  note  must  contain  Si ;  hence,  the  note  must  contain 
as  many  dol's  as  $.9845  is  contained  times  in  the  proceeds. 

$177.21 --$.9845  =  180.     Ans.  $180. 

Pkoop. — Interest  of  $180  for  93  da.  =$2. 79; 
and  $180  — $2. 79  =$177. 21 

Hence,  divide  the  proceeds  of  the  note  hij  the  proceeds  o/Sl  on 
the  same  conditions ;  the  quotient  will  be  the  face  of  the  note, 

2.  For  what  sum  must  a  note  be  made,  at  3  mon.,  so 
that,  when  discounted  at  a  bank  at  6  5^,  the  amount 
received  will  be  $393.80  ?  Ans.  $400. 

3.  I  wish  to  obtain  $500  for  60  days  :  for  what  sum 
must  the  note  be  given?  Ans.  $505.305-1- 

RevvEW. — 241.  How  find  tho  f.ico  of  a  note,  which,  when  discounted, 
the  pryccodd  shall  bo  a  given  sum  ? 


PROFIT    AND   LOSS.  245 

Art.  242.  profit  and   loss 

Are  terms  used  to  express  the  gain  or  loss  in  business. 
In  Profit  and  Loss,  four  quantities  are  considered ; 
Ist,  the  cost  price ;    2d,  the  selling  price ;    3d,  the  amount  of 
gain  or  loss ;   4th,  the  rate  per  cent,  of  gain  or  loss. 

Art.  243.  CASE  I. 

To  find  the  amount  of  profit  or  loss,  when  the  cost 
price  and  rate  per  cent,  of  gain  or  loss  are  given. 

1.  A  merchant  bought  a  piece  of  cloth  for  $40,  and 
sold  it  at  10  %  profit :  how  much  did  he  gain  ? 

Solution, — 10  pr.  ct.  of  $40  is  $4,  the  required  gain. 

2.  A  merchant  bought  a  bale  of  cotton  for  880,  which 
he  sold  at  8  ^  loss  :  what  did  he  lose  ? 

Solution.— 8  pr.  ct.  of  $80  is  $6.40,  the  loss. 

Rule  for  Case  I. — Find  the  given  per  cent,  of  the  cost  price, 
and  the  result  will  be  the  gain  or  loss.     Art.  210. 

Note. — The  rate  per  cent,  of  gain  or  loss  always  refers  to  the 
purchase  or  cost  price,  and  not  to  the  selling  price. 

3.  A  merchant  sold  goods,  that  cost  8150,  and  gained 
10  %  :  what  was  his  gain?  Ans.  815. 

4.  A  peddler  sold  goods,  that  cost  8874,  at  a  gain 
of  25  ^  :  required  his  gain.  Ans.  8218.50 

5.  I  bought  goods  for  8500,  and  sold  them  at  12  % 
profit :  what  sum  did  they  bring  ?  Ans.  8560. 

6.  Sold  goods  that  cost  8382.50,  at  a  loss  of  4  ^  : 
what  sum  did  they  bring  ?  Ans.  8367 .  20 

Art.  244.  CASE  II. 

To  find  the  selling  price,  when  the  cost  price  is 
known,  so  that  a  given  rate  ^  may  bt:  gained  or  lost. 

Review. — 242.  "What  are  profit  and  loss  ?  What  four  quantities  are 
considered  ?  243.  What  is  Case  1  ?  What  the  Eule  ?  Note.  To  what 
does  the  rate  per  cent,  refer  V    244.  What  is  Case  2  ? 


246  RAY'S    PRACTICAL    ARITHMETIC. 

1 .  Tea  costs  60  cents  per  lb. :  at  what  rate  must  it  be 
sold  to  gain  25  %  ? 

Solution.— 25  %  of  60  cts.  is  60  X -25  =  15  cts.,  and  60  cts. 
4- 15  cts.  =  75  cts.  per  lb.     Ans. 

Or,  thus:  25  per  cent.  =  j2jj^y=i,  and  1  of  60  cts.  =  15  cts., 
■which,  added  to  60  cts.,  gives  75  cts.  for  the  selling  price  per  lb. 

If  it  were  required  to  lose  so  much  per  cent.,  the  percentage 
of  the  cost  price  must  be  subtracted  from  it. 

Rule  for  Case  II. — Find  the  percentage  of  the  cost  pHce, 
Art.  210,  and  add  it  to,  or  subtract  it  from,  the  cost,  as  may 
be  required ;  the  result  will  be  the  selling  pi-ice. 

2.  Silk  costs  90  cts.  per  yard :  at  what  price  per  yd. 
must  it  be  sold,  to  gain  25  ^  ?  Ans.  S1.12A 

At  what  price,  to  lose  10  %?  Ans.  §0.81 

3.  If  cloth  cost  S4.37^  per  yard,  at  what  price  per 
yd.  must  it  be  sold,  to  gain  33.J  ^?  Ans.  §5.83-|- 

At  what  price,  to  lose  20  %  ?  Ans.  §3.50 

4.  Cheese  cost  §8.50  per  cwt. :  at  how  much  per  cwt. 
must  it  be  sold,  to  gain  20  ^  ?  Ans.  §10.20 

At  how  much,  to  lose  20  %  ?  Ans.  §6.80 

5.  Bought  40  yards  of  cloth  for  §300  :  at  how  much 
a  yard  must  it  be  sold,  to  gain  20  %  ?  Ans.  §9. 

At  how  much  a  yd.,  to  lose  20  ^  ?  Ans.  §6. 

Art.  245.   of  marking  goods. 

The  rate  per  cent,  of  profit  at  which  merchants  mark 
the  selling  price  of  goods,  is  generally  some  aliquot  part 
of  100.     The  following:  are  those  most  used: 


5    per  cent.  =  r\-^ 

8] =  t"., 

10    =t'd. 


12.'y  per  cent.  =  L 


^"3 —  6' 

20     =A,   50  — » 


25    per  cent.  =  I 


4> 
^^3 =3' 


o' 


Review.— 244.  What  is  tho  Rulo  for  Case  2?  245.  What  part  of 
any  thing  is  5  per  cent.?  8  and  1  third?  10?  12  and  a  half?  16  and 
2  thirds  ?     20  ?     25  ?     83  and  1  third  ?     50  por  cent  ? 


PROFIT   AND   LOSS.  247 

To  mark  goods  for  different  rates  of  profit,  Add  to  the  cost 
price  such  part  of  itself  as  the  rate  per  cent,  is  part  of  100. 

6.  To  make  10  ^  profit,  what  must  calico  be  marked, 
that  cost  lOcts.  per  yard?  15  cts,  ?  20  cts.  ?  30  cts.  ? 
40cts.  ?     50  cts.  ?     GO  cts.  ?  J.«s.  to  last,  66  cts. 

7.  To  make  12A  %  profit,  how  mark  muslin  that  cost 
8  cts.  per  yard.  ?     12  cts.  ?     16  cts.  ?     20  cts.  ? 

Ans.  to  last,  22i  cts. 

8.  To  make  20  %  profit,  how  mark  ribbons  that  cost 
10  cts.  per  yd. ?     15  cts.?    25  cts.  ?     ^?is.  to  last,  30  cts. 

9.  To  make  25  5^  profit,  how  mark  cloth  that  cost  $1 
per  yard?     81.20?     81.50?     82?     83?    84?     ^^'l 

Ans.  to  last,  87.50 

10.  To  make  33^  ^  profit,  how  mark  ginghams  that 
cost  25  cts.  per  yd.?     50  cts.  ?  Ans.  to  last,  ^o^l  cts. 

11.  To  make  50  ^  profit,  how  mark  shawls  that  cost 
82  ?    83  ?    84  ?    85  ?    87  ?  Am.  to  last,  810 .  50 

Art.  246.   case   ill. 

To  find  the  rate  per  cent,  of  profit  or  loss,  when  the 
cost  and  selling  price  are  given. 

1.  I  sold  cloth  at  85  a  yard,  that  cost  84  a  yard:  what 
was  the  gain  %  ? 

Solution. — Take  the  difference  between  the  cost  and  selling 
price,  the  gain  is  $1  on  what  cost  $4;  then  find  what  per  cent.  $1 
is  of  $4;  this,  (Art.  211),  is  \  =  .2b  =  2b  per  cent.     Ans. 

Rule  for  Case  III. —  Take  the  difference  between  the  cost 
price  and  selling  price,  and  find  what  per  cent,  this  is  of  the 
cost  price. 

2.  A  man  paid  875  for  a  horse,  and  sold  him  for  8105 : 
what  %  did  he  gain  ?  Ans.  40  % . 

3.  I  bought  a  piece  of  cloth  for  830,  and  sold  it  for 
840  :  what  was  the  %  profit?  Ans.  33|  %. 

Review. — 245.  How  mark  goods  to  sell  at  different  rates  of  profit? 
246.  What  is  Case  3  ?    What  the  Rule  ? 


248  RAYS   PRACTICAL    ARITHMETIC. 

4.  Clotli  cost  25  cts.  a  yard,  and  sold  for  30  cts.  a  yd. : 
what  the  gain  %  ?  Ans.  20  % . 

5.  Muslin  that  cost  20  cts.  a  yard,  is  sold  at  21  cts.  a 
yard:  what  the  %  profit?  Ans.  5  ^. 

6.  Bought  cloth  at  88  a  yard,  and  sold  it  at  $9  a  yard : 
what  %  profit  did  I  make?  Ans.  12^  %. 

7.  Muslin  that  cost  30  cts.  a  yard,  is  sold  at  24  cts.  a 
yard  :  what  %  is  lost  ?  Ans.  20  %  . 

8.  A  bought  40  bales  of  cotton,  at  840  each,  and  sold 
it  at  a  profit  of  8704  ;  what  did  he  make  ?     Ans.  44  ^. 

Art.  247.  CASE   IV. 

To  find  the  cost  price,  when  the  selling  price  and  rate 
per  cent,  of  profit  and  loss  are  given. 

1.  Cloth  sold  at  85  a  yard,  pays  25  %  profit:  required 
the  cost  price  per  yard. 

Solution. — When  the  cost  is  $1,  and  25  per  cent,  is  gained, 
the  selling  price  is  81.25  (Art.  244);  hence,  as  many  times 
as  81.25  is  contained  in  the  selling  price,  so  many  times  is  81  con- 
tained in  the  cost:  the  cost  price  is  as  many  dollars  as  8l«25  is 
contained  times  in  $5.     85 -r- 81 -25  =  4.     Aris.  ^i. 

Proof.— 25  %  of  $4  is  $1;  and  $4-f  81=$5. 

2.  If,  by  selling  cloth  at  89  a  yard,  10  ^  is  lost,  what 
was  the  cost  price  ? 

Solution. — When  the  cost  is  $1,  and  10  per  cent,  is  lost,  the 
Belling  price  is  8  •  90  (Art.  244);  hence,  as  often  as  $.90  is  con- 
tained in  the  selling  price,  so  many  times  is  81  contained  in  the 
purchase  price:  the  cost  is  as  many  dollars  as  8.00  is  contained 
times  in  89.     S9  -^  8  •  90  =  10.     Ans.  810. 

Proof.— 10  fo  of  810  is  $1 ;  and  $10  — $1  =$9,  the  selling 
price. 

*3.  Cloth  sold  at  8G  a  yard,  pays  20  %  profit :  what 
was  the  cost  price  per  yd.?  Ans.  85. 

Review.- 247.  What  is  Caso  4?     Kulo?     Kule  by  proix>rtion? 


PROFIT  AND  LOSS.  249 

*4.  By  selling  cloth  at  S3  a  yard,  25  ^  was  lost : 
wliat  was  the  cost  price  per  yd.?  Ans.  S4. 

Rule  for  Case  IV. — Add  to,  or  subtract  from,  $1,  as  may 
be  required^  the  per  cent,  of  gain  or  loss  on  ^1 ;  divide  the 
selling  price  by  the  result ;   the  quotient  will  be  the  cost. 

Or,  by  Proportion.  As  100  increased  by  the  per  cent,  of 
gain,  or  diminished  by  the  per  cent,  of  loss,  is  to  100,  so  is  the 
selling  price  to  the  cost. 

Observe,  that  the  per  cent,  of  profit  or  loss  is  always  calculated 
on  the  cost,  and  not  on  the  selling  price. 

5.  A  jockey  sold  a  horse  for  S75,  and  gained  25  ^  : 
what  did  the  horse  cost  him?  Ans.  $Q0. 

6.  A  .jockey  sold  a  horse  for  S75,  and  lost  25  ^  :  what 
did  the  horse  cost  him?  Ans.  $100. 

7.  A  grocer,  by  selling  coffee  at  22  cts.  a  lb.,  gains 
10  ^  :  find  the  purchase  price  per  lb.  Ans.  20  cts. 

8.  By  selling  cloth  at  $8.10  per  yard,  I  gain  12^^-  ^  : 
what  the  purchase  price  per  yd.  ?  Ans.  $7 .20 

9.  By  selling  tea  at  $1.19  per  pound,  I  lost  15  ^: 
what  the  cost  price  per  lb.?  Ans.  $1.40 


Art.  248.   PROMISCUOUS    EXAMPLES. 

1.  A  buys  30  yd.  muslin,  at  6  cts.  a  yd.,  and  sells  it 
at  25  %  profit :  what  does  he  gain  ?  Ans.  45  cts. 

2.  B  bought  40  yd.  of  cloth  for  $250  :  at  what  price 
per  yd.  must  he  sell  to  gain  20  %  ?  Ans.  $7.50 

3.  A  bought  1  hhd.  of  wine  for  $75,  and  sold  it  for 
40  cts.  a  qt. :  what  the  %  profit  ?  Ans.  34|  % . 

4.  The  population  of  New  York,  in  1830,  was  1918604; 
in  1840  it  was  2428921  :  find  the  gain  per  cent,  in 
10  years.  Ans.  26.5  %  + 

5.  The  population  of  Ohio,  in  1830,  was  937903 ;  in 
1840  it  was  1519467:  what  was  the  gain  per  cent, 
in  10  years?  Ans.  62  ^  4~ 


250  RAY'S  PRACTICAL  ARITHMETIC. 

6.  If  calico  is  sold  at  42  cts.  a  yd.,  at  a  gain  of  10  %^ 
what  was  the  cost  price  ?  what  the  gain  ^,  if  sold  at  51 
cts.  a  yard?        Ans.  Cost,  38j^jCts.  a  yd.    Gain,  33|  ^. 

7.  Sold  a  bushel  of  rye  for  SI,  and  gained  25  ^  ; 
purchased  a  bu.  of  wheat  with  the  §1,  and  sold  it  at  a 
loss  of  25  ^  :  what  did  I  lose  ?  Ans.  5  cts. 

8.  A  merchant  bought  1-1  pieces  of  cloth  at  89.60  each; 
sold  5  pieces  at  S11.40  each,  and  4  pieces  at  812  each: 
at  how  much  a  piece  must  he  sell  the  remainder,  to 
gain  20  %  on  the  whole?  Ans.  $8,256 

9.  Sold  wine  at  81.29  a  gal.,  and  lost  14  ^  :  at  what 
price  per  gal.  must  it  sell,  to  gain  14  ^  ?       Ans.  81.71 

10.  Sold  cloth  at  81.36  a  yd.,  and  lost  15  %  :  what  % 
would  I  gain  by  selling  at  81 .856  a  yd.?       Ans.  IQ  %. 

11.  Sold  silk  at  81.96  a  yd.,  and  gained  12  ^  :  at  what 
price  per  yd.  should  I  sell,  to  lose  16  %  ?      Ans.  81.47 

12.  Sold  satin  at  81.682  a  yd.,  and  gained  16  ^  :  what 
%  will  I  lose,  by  selling  at  81.247  a  yd.?     Ans.  14  %. 

13.  How  much  cloth,  at  85  a  yd.,  must  I  buy,  to  clear 
8100  by  selling  it  at  25  %  profit  ?  Ans.  80  yd. 

14.  A  grocer  buys  200  casks  of  raisins  at  82.50  per 
cask  ;  by  selling  at  5  cts.  per  pound,  he  gains  20  ^  : 
what  was  the  weight  of  each  cask  ?  Ans.  60  lb. 

15.  Sold  a  quantity  of  corn,  at  81  per  bu.,  and  gained  25  ^  ; 
sold  of  the  same  to  the  amount  of  859.40,  and  gained  35  %  : 
at  what  rate  did  I  sell :  how  many  bu.  in  the  last  lot  ? 

Ans.  81 .  08  per  bu.,  and  55  bu. 
C^*  For  additional  problems,  see  Ray's  Test  Examples. 

ASSESSMENT    OF    TAXES. 

Art.  249.  A  tax,  is  a  sura  assessed  on  the  citizens  of 
a  town,  county,  state,  or  district,  for  public  purposes. 

Taxes  are  of  two  kinds — a  j^roperty  tax  and  a  poll-tax. 

A  PROPERTY  TAX  IS  a  Certain  per  cent  assessed  on  the  taxable 
property  held  by  each  person. 


ASSESSMENT  OF   TAXES.  251 

A  POLL-TAX  is  a  specific  sum  assessed  on  male  citizens  over  21 
years  of  age.     Each  person  so  taxed  is  called  a  poll. 

PiEM. — In  some  States  the  whole  tax  is  raised  on  property ;  in 
other  States,  partly  on  property  and  partly  on  polls. 

Art.  250.  When  a  tax  is  to  be  assessed,  first  obtain 
a  list  or  inventory  of  the  amount  of  taxable  property, 
from  which  the  tax  is  to  be  collected. 

If  there  be  a  poll-tax,  make  a  list  of  the  polls. 

To  Assess  a  Tax,  Observe  this 

B,ulc. — 1.  If  there  be  a  poll-tax,  Jind  its  amount,  by  multi- 
plying the  tax  on  each  poll  by  the  number  of  polls.  Subtract 
this  from  the  ichole  amount  of  tax  to  be  raised ;  the  remainder 
will  be  the  sum  to  be  raised  07i  property. 

2.  Divide  the  tax  to  be  raised  on  property  by  the  ichole  amount 
of  property ;  the  quotient  will  be  the  per  cent,  of  tax  on  $1. 

3.  Multiply  the  per  cent,  of  tax  on  $1  by  the  amount  of  each 
person's  p)rop)erty,  the  product  icill  be  his  property  tax. 

4.  Add  the  poll-tax,  if  any,  of  each  person  to  his  property 
tax ;  the  sum  will  be  his  whole  tax. 

1.  A  tax  of  S500  is  assessed  in  a  district,  to  build  a 
school-house ;  the  property  is  valued  at  S125000  :  what 
the  ^  of  tax?  Ans.  .00-i,  or  4  m.  on  §1. 

What  the  tax  on  S1650?  Ans.  $6.60 

2.  A  tax  of  89057.60  is  assessed  in  a  county  whose 
taxable  property  is  valued  at  8534650 ;  also,  a  list  of 
1258  polls,  each  taxed  81.25  :  what  the  %  of  tax  on 
property?  Ans.  .014,  or  le.  4m.  on  81. 

Note. — In  preparing  tax  lists,  after  finding  the  per  cent,  of  tax, 
assessors  make  a  table  embracing  the  tax  on  dollars  from  1  to  10; 
then  on  10,  20,  &c.,  to  $100;   then  on  100,  200,  &c.,  to  $1000. 

The  following  table,  computed  for  the  preceding  example,  is  cal- 
culated by  multiplying  the  tax  on  §1  by  the  number  of  dollars  on 
which  the  tax  is  required. 

Keview. — 249.  What  is  a  tax?    What  a  property  tax  ?    A  poll-tax? 
250.  When  a  tax  is  to  be  assessed,  what  is  to  be  first  obtained  ? 
250.  How  assess  a  tax,  Rule  ?    Note.  How  calculate  a  tax-table  ? 


252 


RAY'S    PRACTICAL    ARITHMETIC. 


Tax  Table. — Rate,  14  Mills  on  $1. 


$1 

pays  $.014 

$20  pays 

$.280 

$3U0  pays 

$4,200 

2 

028 

30 

.420 

400 

5.600 

3 

042 

40 

.560 

500 

7.000 

4 

056 

50 

.700 

600 

8.400 

5 

070 

60 

.840 

700 

9.800 

6 

084 

70 

.980 

800 

11.200 

7 

098 

80 

1.120 

900 

12.600 

8 

112 

90 

1.260 

1000 

14.000 

9 

126 

100 

1.400 

2000 

28.000 

10 

140 

200 

2.800 

3000 

42.000 

3.  What,  by  the  Table,  was  A's  tax  ;  his  property  valued 
at  $756,  he  paying  for  2  polls  ? 

OPEEATION. 

$700  is  taxed  $9,800 

This  operation  is  the  the  same          5Q  .700 

as    multiplication,    Art.    82,    and             g  ^  Qg^. 

consists    in    taking    the    product       

of  $.014  by  700,  by  50,  and  by  6,     $756        ....  $10,584 

and  then  finding  their  sum.                     2  polls  pay  2.500 

Ans.  whole  tax,  $13,084 

4.  What,  by  the  Table,  was  B's  tax ;  his  property  valued 
at  $1243,  he  paying  for  3  polls  ?  Ans.  $21 .152 

5.  C's  property  is  valued  at  $3589,  and  he  pays  foi 
4  polls  :  what  is  his  tax?  Ans.  $55,246 

AMERICAN    DUTIES. 

Art.  251.  Duty  is  a  tax,  levied  by  the  Government 
on  goods  imported  from  a  foreign  country. 

Note. — Duties  are  of  two  kinds,  Specific  and  Ad  valorem. 

A  Specific  duty  is  a  fixed  sum  per  tun,  gallon,  yard,  &c.,  without 
regard  to  value. 

An  Ad  valorem  duty,  {according  to  value),  is  a  certain  per  cent, 
of  the  cost  of  the  goods  in  the  country  from  which  they  were 
imported. 

In  reckoning  duties,  deductions  are  made  from  the  gross  weight 
or  measure.     These  are  termed,  draft,  tare,  and  leakage. 


Review.— 251.  "What  is  duty  ?    Specific  duty  ?    Ad  valorem? 


DUTIES.— PARTNERSHIP.  253 

Draft  is  made,  that  the  quantity  may  hold  out  when  retailed. 

On  each  parcel  weighing  112  lb.,  or  less,  the  draft  is  1  lb. 
From  112  lb.  to  224  lb.,  the  draft  is  2  lb. 
From  224  lb.  to  336  lb.,  the  draft  is  3  lb. 
From  336  lb.  to  1120  lb.,  the  draft  is  4  lb. 
From  1120  lb.  to  2016  lb.,  the  draft  is  7  lb. 
Over   2016  lb.  the  draft  is  9  lb. 

Tare  is  an  allowance  (after  deducting  the  draft),  for  the  weight 
of  the  box,  cask,  &c.,  containing  the  goods. 

Gross  weight  is  the  weight  before  deducting  draft  and  tare. 

Net  weight  is  the  weight  after  deducting  the  draft  and  tare. 

Leakage  is  an  allowance  of  2  per  cent,  on  all  liquors  in  caskg, 
paying  duty  by  the  gallon. 

Duties  are  computed  on  what  remains  after  deducting  all  allow- 
•znces.     The  calculations  are  an  application  of  percentage. 

1.  Find  the  duty  on  3  boxes  of  Sugar,  of  1001b., 
182  lb.,  and  264:  lb.,  at  2  cts.  a  lb.,  allowing  for  draft, 
and  deducting  15  ^  for  tare.  Ans.  89.18 

First  deduct  the  draft  (61b.),  then  15  per  cent,  of  the  remainder 
for  tare,  and  compute  the  duty  on  the  last  remainder. 

2.  What  the  duty,  at  20  %  ad  valorem,  on  40  bales  of 
wool,  of  400  lb.  each,  cost,  in  Spain,  25  cts.  a  lb.,  the 
tare  5%?  Ans.  $752.40 

3.  A  merchant  imports  75  eases  of  indigo,  gross 
weight  1961b.  each:  allowing  15  ^  for  tare,  what  the 
duty  at  5  cts.  per  lb.?  Ans.  $618,375 


XVII.    PARTNERSHIP. 

Art.  252.  Partnership  is  an  association  of  persons 
for  the  transaction  of  business  :  such,  is  called  a  Jirm 
or  house;  and  each  member,  a,  partner. 

Review. — 251.  What   are  the  allowances  for  draft?     What  is  tare? 
Gross  weight?     Net  weight?     Leakage?     On  what  are  duties  computed  ? 


254  RAY'S   PRACTICAL    ARITHMETIC. 

The  CAPITAL,  or  stock,  is  the  amount  of  money  or  property 
contributed  by  the  firm. 

The  DIVIDEND  is  the  gain  or  loss  shared  among  the  partners. 

1.  A  and  B.  engaged  in  trade:  A's  capital  was  S200 ; 
B's,  $300;  they  gained  §100:  find  each  partner's  share. 

Solution.— The  whole  c^^pital  is  $200 -{-$300  =  6500.  Of  this 
A  owns  §^§="f>  and,  therefore,  he  should  have  g  of  the  gain: 
B  owns  |ggr=|  of  the  capital,  and  should  have  |  of  the  gain. 

Hence,  A's  gain  will  be  |  of  $100  =  $40.  |    ^^^ 
B's  gain  will  be  §  of  $100  =  $60.  j 

*2.  A  and  B  form  a  partnership,  with  a  capital  of 
S800:  A's  part  is  8300;  B's,  $500;  they  gain  $232: 
what  the  share  of  each?  Ans.  A's,  $87;  B's,  $145. 


TO   FIND   EACH   PARTNER  S   SHARE 
Of  the  gain  or  loss,  when  each  one's  capital  is  used  the  same  time. 

Hule. —  Take  suck  part  of  the  whole  gain  or  loss,  as  each 
partner  s  stock  is  part  of  the  whole  stock. 

Or,  by  Proportion.  As  the  whole  stock  is  to  each  partner  s 
stocky  so  is  the  whole  gain  or  loss  to  each  partner  s  gain  or  loss. 

Proof. — Add  together  the  several  shares;  if  the  work  is 
correct,  the  sum  will  equal  the  whole  gain  or  loss. 

Rem. — 1,  This  rule  is  applicable,  when  required  to  divide  a  sura 
into  parts  having  a  given  ratio  to  each  other;  as  in  Bankruptcy, 
General  Average,  &c. 

2.  Partnerships  are  generally  governed  by  special  agreements, 
which  specify  the  method  of  dividing  gains  or  losses. 

3.  A's  stock  was  $70;  B's,  $150;  C's,  $80;  they  gained 
$120:  what  was  each  man's  share  of  it? 

Ans.  A's,  $28;  B's,  $G0;  C's,  $32. 

4.  A,  B,  and  C  traded  totrether:  A  put  in  $200; 
B,  $400;  C,  $600:  they  gained  $427. 2G:  find  each  man's 
share.       ^ws.  A's,  $71 .21 ;  B's,  $142.42;   C's.  $213.63 

What  was  the  gain  ^^  ?  Ans.  35.005  %. 


PARTNERSHIP.  255 

5.  Divide  890  among  3  persons,  so  that  tlie  parts  shall 
be  to  each  other  as  1,  3,  and  5.  Ans.  $10,  830,  and  850. 

6.  Divide  8735.93  among  4 men,  in  the  ratio  of  2,  3, 
5,  and  7.       ^tzs.  886.58;  8129.87;   8216.45;  8303.03 

7.  A  person  left  an  estate  of  822361,  to  be  divided 
among  6  children,  in  the  ratio  of  their  ages,  which  are 
3,  6,  9,  11,  13,  and  17  yr.:  what  are  the  shares? 

^Tis.  81137;  82274;  83411;  84169;  84927;  86443. 

8.  Divide  8692.23  into  3  parts,  that  shall  be  to  each 
other  as  i,  |,  and  |.    ^rzs.  8127.60;  8229.68;  8334.95 

Art.  253.    OF    BANKRUPTCY. 
A  BANKRUPT  is  onc  who  fails  in  business. 

9.  A  man,  failing,  owes  A  8175  ;  B,  8500 ;  C,  8600 ; 
D,  8210;  E,  842.50;  F,  820;  G,  810;  his  property  is 
worth  8934.50:  what  will  be  each  creditor's  share? 

Ans.  A's,  8105;     C's,  8360;     E's,  825.50; 

B's,  8300;     D's,  8126;     F's,  812.00;  G's,  86. 

Note. — Such  questions  may  also  be  solved  by  finding  what  can 
be  paid  on  §1,  and  multiplying  this  by  each  creditor's  claim. 

10.  A  man  owes  A  8234;  B,  8175;  C,  8326:  his 
property  is  worth  8492.45:  what  can  he  pay  on  81;  and 
what  will  each  creditor  get?  Ans.  67  cts.  on  81; 

A,  8156.78;  B,  8117.25;  C,  8218.42 

Art.  254.    GENERAL  AVERAGE 

Is  the  method  of  apportioning  among  the  owners  of  a 
ship  and  cargo,  losses  occasioned  by  casualties  at  sea. 

11.  A,  B,  and  C  freighted  a  ship  with  108  tuns  of  wine. 
A  owned  48,  B  36,  and  C  24  tuns;  they  were  obliged 
to  cast  45  tuns  overboard :  how  much  of  the  loss  must 
each  sustain?  Ans.  A,  20;   B,  15;  C,  10  tuns. 


Review. — 252.  What  is  partnership?  A  firm?  A  partner?  The 
Capital?  Dividend?  How  find  each  partner's  share  of  the  gain  or  loss, 
Rule  1    How,  by  proportion  ?    Rem. — To  what  is  this  rule  applicable  ? 


256  RAY'S    PRACTICAL    ARITHMETIC. 

12.  From  a  ship  valued  at  810000,  with  a  cargo  valued 
at  815000,  there  was  thrown  overboard  goods  valued  at 
81125  :  what  ^  was  the  general  average,  and  what  was 
the  loss  of  A,  whose  goods  were  valued  at  §2150  ? 

Ans.  General  average,  4^  ^  ;  A's  loss,  896.75 

What  the  captain's  loss,  he  owning  |  of  the  ship  ? 

Ans.  8168.75 

Art.  255.   PAETNERSHIP    WITH    TIME. 

1.  A  and  B  built  a  wall  for  882  ;  A  had  4  men  at 
work  5  days,  and  B,  3  men  7  days  :  how  should  they 
divide  the  money  ? 

SoLU. — The  work  of  4  men  5  da.  equals  the  work  of  4X5, 
or  20  men  1  da.;  and  the  work  of  3  men  7  da.,  equals  the  work 
of  3X7,  or  21  men  1  da.:  it  is  then  required  to  divide  S82  into 
two  parts  having  the  same  ratio  to  each  other  as  20  to  21. 

Hence,     A's  part  is  2Q  of  $82=S40.  | 

B's  part  is  |1  of  $82  =  $42.  |  ^^^' 

2.  A  put  in  trade  850  for  4mon.  ;  B,  860  for  5  mon. ; 
they  gained  824 :  what  is  each  man's  share  ? 

Solution. — $50  for  4  mon.  equals  $50X4  =  $200  for  1  mon,; 
and  $60  for  5  mon.  equals  $00X5  =  $300  for  1  mon.  Hence, 
divide  $24  into  two  parts  having  the  same  ratio  as  200  to  300. 

).60') 

500  —  5  "  •*- >'-i.40  J 

Hence,  to  find  each  partner's  share  of  the  gain  or  loss,  when 
iime  is  regarded, 

Multiply  each  partner  s  stock  by  the  time  it  was  employed ; 
then  take  such  part  of  the  gain  or  loss  as  each  partner  s  product 
is  part  of  the  sum  of  all  the  products. 

Or,  by  Proportiox.  Multiply  each  partner  s  stock  by  the  time 
employed;  then,  as  the  sum  of  the  products  is  to  each  partner  s 
product^  so  is  the  whole  gain  or  loss  to  each  partner  s  share. 

TIkvikw. — 253.  What  is  a  bankrupt  ?  Note.  How  may  questions  in 
biiiikrui)tcy  1)0  solved?     2'A.  What  is  general  average? 

255.  When  timo  is  regarded  in  partnership,  how  find  each  partner's  share  ? 


This,  (Art.  252),  gives  A  fgg  =  f  of  $24=$  9.60 
and  B  lofi^lof  $24=$14.-"  ^ 


EQUATION  OF  PAYMENTS.  257 

3.  A  and  B  hire  a  pasture  for  $54  :  A  pastures  23 
horses  27  da. ;    B,  21  horses  39  da.  :  what  will  each  pay? 

Ans.  A,  823. 28|;  B,  830.711 

4.  A  put  in  8300  for  5  mon. ;  B,  8400  for  8  mon. ; 
C,  8500  for  3  mon.  :  they  lost  8100  :  find  each  one's  loss. 

Ans.  A's,  824.19J]  ;  B's,  851. 61/^;  C's,  824.191} 

5.  A,  B,  and  C  hire  a  pasture  for  $18.12  :  A  pastures 
6  cows  30  da. ;  B,  5  cows  40  da. ;  C,  8  cows  28  da. :  what 
shall  each  pay?  Ans.  A,  85.40;  B,  m  ;  C,  86.72 

6.  Three  men  formed  a  partnership  for  16  mon.  :  A 
put  in  at  first  8300,  and  at  the  end  of  8  mon.,  8100 
more  ;  B  put  in  at  first  8600.  but,  at  the  end  of  10  mon., 
drew  out  8300  ;  C  put  in  at  first  8500,  and,  at  the  end 
of  12  mon.,  8400  more ;  they  gained  8759  :  find  each 
man's  share. 

A71S.  A's,  8184.80;  B's,  8257.40;  C's,  8316.80 

7.  A  and  B  are  partners  :  A  put  in  8800  for  12  mon., 
and  B,  8500.  What  sum  must  B  put  in  at  the  end  of  7 
mon.,  to  entitle  him  to  half  the  yr.'s  profits?   Ans.  8720. 

g^^  For  additional  problems,  see  Ray's  Test  Examples. 


XYIII.    EQUATION   OF   PAYMENTS. 

Art.  256.  Equation  or  equality  of  Payments  is  the 
method  of  finding  the  mean  or  average  time  of  making 
two  or  more  payments,  due  at  different  times. 

The  rule  for  finding  the  mean  or  equated  time,  is  based  on 
the  principle,  that 

The  interest  of  any  sum  for  any  given  period,  is  equal  to  the 
Int.  of  half  the  sum  for  twice  the  period;  of  one-third  of  the 
sum  for  three  times  the  period,  and  so  on.     Thus, 

The  Int.  of  $2  for  1  mon.  =  Int.  of  $1  for    2  mon. 
Int.  of  $4  for  5  mon.  =  Int.  of  $1  for  20  mon. 


Review. — 256.  "What  is  Equation  of  Payments  ?     On  what  principle  is 
the  rule  for  finding  equated  time  based  ?     Give  examples. 
3d    Bk.  17 


258  RAY'S   PRACTICAL  ARITHMETIC. 

Example. — The  Int.  of  $4  for  5  mon.,  at  6  per  cent.,  is  10  cents 
(Art.  224) :  the  Int.  of  $1  for  20  mon.,  is  also  10  cents  (Art.  223). 

Art.  257.  1.  A  owes  B  $2  due  in  3  mon.,  and  $4  due 
in  6  mon. :  at  what  period  can  both  sums  be  paid,  neither 
party  being  the  loser? 

Prom  Art.  255,  it  follows,  that, 
Int.  of  $2  for  3  mon.  =  int.  of  $1  for  2  X  3  =    6  mon. 
Int  of  $4  for  6  mon.  =  int.  of  $1  for  4  X  6  =  24  mon. 

$6  for  -  mon.  =  int.  of  $1  for  30  mon. 

Now  find  in  what  time  $6  will  produce  the  same  Int.  as  §1 
in  30  mon.  At  $6  is  6  times  $1,  it  will  produce  the  same  Int.  in  h  of 
.the  time  (Art.  256);  that  is,  in  30  mon.-T-6  =  5  mon.  Ans. 

Proof. — Int.  of  $2  for  3  mon.,  at6^,=2Xl^=  3  cts. 
Int.  of  $4  for  6  mon.,  .  .  .  =4  X  3  =  12  cts. 
Int.  of  $6  for  5  mon.,  .     .    .  =  6  X  2^  =  15  cts. 

*2.  A  owes  B  S2  due  in  4  mon.,  and  $6  due  in  8  mon. : 
find  the  average  time  of  paying  both  sums.     Ans.  7  mon. 

COMMON   RULE   FOR   EQUATION   OF   PAYMENTS. 

Multiply  each  payment  by  the  time  to  elapse  till  it  becomes 
due ;  divide  the  sum  of  the  products  by  the  sum  of  the  payments; 
the  quotient  will  be  the  equated  time. 

"When  one  of  the  payments  is  due  on  the  day  from  which  the 
equated  time  is  reckoned,  its  product  is  0;  but,  in  finding  the  sum 
of  the  payments,  this  must  be  added  with  the  others.    See  Ex.  6. 

3.  A  owes  B  $8,  due  in  5  mon.,  and  84  due  in  8  mon. : 
find  the  mean  time  of  payment.  .4ns.  6  mon. 

4.  A  buys  SI 500  worth  of  goods;  S250  are  to  be  paid 
in  2  mon.;  $500  in  5  mon.;  $750  in  8  mon.:  find  the 
mean  time  of  payment.  Ans.  6  mon. 

5.  A  owes  B  S300;  1  third  due  in  6  mon.;   1   fourth 

Keview. — 257.  What  is  tho  common  rule  for  Equation  of  Payments? 
"When  one  of  tho  sums  is  to  bo  paid  down,  how  proceed  ? 


EQUATION  OF   PAYMENTS.  259 

in  8  raon. ;    the  remainder  in  12  mon. :  what  the  average 
time  of  payment  ?  Aiis.  9  mon. 

6.  I  buy  $200  worth  of  goods ;  1  fifth  to  be  paid  now ; 
2  fifths  in  5  mon. ;  the  rest  in  10  mon. :  what  the  average 
time  of  paying  all  ?  Ans.  6  mon. 

Art.  258.  In  finding  the  Average  or  3Iean  time  for 
the  payment  of  several  sums  due  at  difierent  times,  any 
date  may  be  taken  from  which  to  reckon  the  time. 

7.  A  merchant  buys  goods  as  follows,  on  60  days 
credit:  May  1st,  1848,  $100;  June  15th,  $200:  what 
the  average  time  of  payment?  Ans.  July  30th. 

Counting  from  May  Ist,  it  is  60  days  to  the  first  payment,  and 
105  days  to  the  second. 

$100  X    60=   6000  27000-^300  =  90. 

$200  X  105  =  21000       ^^^^  go  ^^yg  ^^^^  ^j^y  ^g^^  ^j^^^  .g^ 

$300  27000  July  30th. 

Assuming  April  1st  as  the  day  from  which  to  count,  the  period 
is  120  days,  which  makes  the  same  day  of  payment. 

8.  I  bought  goods  on  90  days  credit,  as  follows :  April 
2d,  1853,  $200^;  June  1st,  $300  :  what  the  average  time 
of  payment?  Ans.  Aug.  6th. 

Art.  259.  The  preceding  rule,  generally  used,  sup- 
poses discount  and  interest  paid  in  advance  to  be  equal ; 
but  this  (Art.  239,  Rem.  2)  is  not  correct. 

The  following,  based  on  true  discount  (Art.  235),  is  the 

TRUE  RULE  FOR  THE  EQUATION  OF  PAYMENTS. 

Find  the  present  worth  of  each  debt  (Art.  237),  thenjind  the 
TIME  (Art.  230),  at  which  the  sum  of  the  present  worths  will 
amount  to  the  sum  of  the  debts  :  this  gives  the  true  equated  time. 

9.  A  owes  $103  due  in  6  mon.,  and  $106  due  in  12 
mon.:  find  the  true  mean  time  of  payment.  Ans.  9 mon. 

Review. — 258.  To  find  the  mean  timo,  from  what  date  do  you  reckon  ? 
259.  Is  the  common  rule  for  Equation  of  Payments  strictly  accurate  ? 
What  is  the  true  rule  for  Equation  of  Payments? 


260  RAY'S  PRACTICAL   ARITHMETIC. 

XIX.    ALLIGATION   MEDIAL. 

Art.  260.  Alligation  medial  is  the  method  of  finding 
the  mean  or  average  price  of  a  mixture,  when  the  ingre- 
dients composing  it,  and  their  prices,  are  known. 

1.  I  mix  4  pounds  of  tea,  worth  40  cts.  a  lb.,  with  6  lb. 
worth  50  cts.  a  lb. :  what  is  1  lb.  of  the  mixture  worth  ? 

Solution. — 4  lb.  at  40  cts.  opkration. 

per  lb.=$1.60,  and  61b.  at  4  lb.  at  40  cts.  cost  SI. 60' 

50cts.=$3.00;   making  the  6  lb.  at  50  cts.  cost     3.00 
total  value   of  the   10  lb.= 


$4.60:    hence,  lib.  cost  J^ 


10  lb.  84.60 


of  $4.60,  or  46  cts.  84.60^10  =  46  cts.,  cost  ofl  lb. 

*2.  Mix  6  lb.  sugar,  at  3  cts.  a  lb.,  with  4  lb.  at  8  cts.  a 
lb.,  what  will  1  lb.  of  the  mixture  be  worth?    Am.  5 cts. 

Kule. — Divide  the  lohole  cost  by  the  whole  number  of  ingre- 
dients;  the  quotient  will  he  the  average  or  mean  2)r ice. 

Note. — The  principles  of  this  rule  may  be  applied  to  the  solu- 
tion of  many  examples  not  embraced  in  the  definition. 

3.  Mix  251b.  sugar  at  12  cts.  a  lb.,  251b.  at  18  cts., 
and  40  lb.  at  25  cts. :  what  is  1  lb.  of  the  mixture  worth  ? 

Ans.  19^  cts. 

4.  A  mixes  3  gal.  water,  with  12  gal.  wine,  at  50  cts.  a 
gal.  :  what  is  1  gal.  of  the  mixture  worth  ?    Ans.  40  cts. 

5.  I  have  30  sheep:  10  are  worth  $3  each;  12,  $4  each; 
the  rest,  $9  each  :  find  the  average  value.  Ans.  $5. 

6.  On  a  certain  day  the  mercury  in  the  thermometer 
stood  as  follows:  from  6  till  10  A.  M.  at  63°;  from  10 
A.  M.  till  1  P.  M.,  70°;  from  1  till  3  P.  M.,  75°;  from 
3  till  7  P.  M.,  73°;  from  7  P.  M.  till  6  A.  M.  of  the  next 
day,  55°.  What  was  the  mean  temperature  of  the  day, 
from  sunrise  to  sunrise?  Ans.  621.° 

Multiply  the  number  of  hours  by  their  mean  temperature; 
divide  the  sum  of  the  products  by  24,  the  sum  of  the  hours. 

Kevikw.— 2(10.  What  is  Alligation  Medial  ?    What  tho  Kulo? 


XX.    ANALYSIS. 

Art.  261.  Analysis  is  the  separation  of  things  into 
their  elements  or  parts.  In  Arithmetic,  it  is  the  method 
of  solution  by  reasoning  according  to  the  nature  of  the 
question,  without  reference  to  special  rules. 

EXAMPLES   FOR   MENTAL   SOLUTION. 

1.  If  5  oranges  cost  15  cts.,  what  cost  4  oranges? 

Analysis. — 1  orange  is  ^  of  5  oranges,  and  will  cost  4  as 
much ;  i  of  15  cents  is  3  cents,  the  cost  of  1  orange ;  4  oranges 
"will  cost  4  times  as  much  as  1  orange ;  4  times  3  cts.=  12  cts.  Ans. 

Here,  we  first  find  the  price  of  owe,  as  it  is  easier  to  compute 
from  the  value  of  one,  than  from  that  of  any  other  number. 

2.  If  5  sheep  cost  S20,  what  will  9  sheep  cost  ? 

3.  If  5  bl.  flour  cost  840,  what  will  3  bl.  cost? 

4.  If  3  lemons  cost  12  cts.,  how  many  will  28  cts.  buy  ? 

Analysis. — 1  lemon  is  I  of  3  lemons,  and  will  cost  i  as  much; 
but  i  of  12  cents  is  4  cents,  the  cost  of  1  lemon.  If  4  cents 
buy  1  lemon,  28  cents  will  buy  as  many  lemons  as  4  cents  are 
contained  times  in  28  cents  ;  that  is,  7.     Ans.  7  lemons. 

5.  If  5  barrels  of  flour  cost  $15,  how  many  barrels 
of  flour  can  be  purchased  for  $21  ? 

6.  If  6  lb.  of  sugar  cost  30  cts.,  how  many  pounds  of 
sugar  can  be  bought  for  50  cts.  ? 

7.  If  7  yards  of  cloth  cost  828,  how  many  yards  of 
cloth  will  840  buy? 

8.  James  had  28  cts.,  and  spent  |  of  them  for  oranges 
at  2  cts.  each  :  how  many  oranges  did  he  buy  ? 

Analysis. — ^  of  28  is  4,  and  |  are  3  times  4  =  12.  If  2  cents 
buy  1  orange,  12  cents  will  buy  as  many  oranges  as  2  cents  are 
contained  times  in  12  cents,  that  is  G.     Ans.  G  oranges. 

9.  A  man  having  840,  spent  |  of  it  for  cloth  at  $2  a 
yard  :  how  many  yards  did  he  purchase? 

10.   I  of  48  are  how  many  times  10? 


Eeview. — 2G1.  What  id  Analysis  ?    What  is  it  in  Arithmetic  ? 

261 


26-2  RAY'S    PRACTICAL   ARITHMETIC. 

11.  5  of  45  are  how  many  times  8? 

12.  1  sold  a  watch  for  818,  which  was  |  of  its  cost : 
how  much  did  it  cost,  and  what  did  I  lose  ? 

Anal.— Since  $18  is  3  fifths  of  the  cost,  i  of  18  will  be  1  fifth; 
and  I  of  18  is  6.     If  6  is  1  fifth,  6  fifths  will  be  5XG=30. 
Hence,  the  cost  is  $30,  and  $30— $18=$12,  the  loss. 

Here,  we  reason  from  several  parts  to  one  part ;  then  from 
one  part  to  the  required  number. 

Rem, — To  avoid  difficulty  in  analyzing  examples  of  this  kind, 
observe  that  the  numerator  of  the  fraction  shows  the  number  of 
parts  taken.  Art.  124.  Thus,  in  the  preceding  example,  18  is  three 
parts  (fifths) ;  and  1  part  (fifth)  is  1  third  of  3  parts  (fifths). 

13.  8  is  I  of  what  number  ?         12  is  |  of  what  ? 

14.  16  is  I  of  what  aumber?         15  is  |  of  what? 

15.  A  farmer  sold  a  wagon  for  S45,  which  was  ^  of  its 
cost ;  he  paid  for  it  with  sheep  at  $3  a  head  :  how  many 
sheep  were  required  ?  Ans.  21. 

16.  25  is   g-   of  how  many  times  4? 

17.  21  is  -^j  of  how  many  times  4  ? 

18.  I  of  40  is  I  of  what  number  ? 

Analysis.—^  of  40  is  10,  and  |  is  3  times  10=30.  If  30 
is  ^,  ^  of  30  =  G,  is  ^ ;  and  ^,  or  the  required  number,  are  7X6=42. 

19.  I  of  12  is  4  of  what  number? 

20.  I  of  27  is  f  of  what  number  ? 

21.  If  J  bushel  barley  cost  20cts.,  what  cost  }  bu.  ? 

Analysis. — If  ^  cost  20cts.,  1  bu.  will  cost  3  times  as  much, 
or  GOcts. ;  and  i  bu.  will  cost  X  of  this,  which  is  15  cts.  Ans. 

22.  If  I  yd.  muslin  cost  24 cts.,  what  cost  |  yd.? 

Analysis. — 1  yd.  will  cost  \  as  much  as  | ;  1  of  24  cts.  is  12  cts., 
the  cost  of  1  of  a  yd. ;  hence,  1  yd.  will  cost  3  times  12  cts., 
or  3Gcts.  If  1  yd.  costs  36  cts.,  X  will  cost  9  cts.,  and  |  three 
times  9  cts.,  which  are  27  cts.  Ana. 

Here,  wo  reason  from  several  parts  to  one  part;  then,  from 
one  part  to  the  whole ;  then,  from  the  whole  to  a  part ;  lastly, 
from  one  part  to  several  parts. 


ANALYSIS.  263 

23.  If  \  yd.  silk  cost  20  cts.,  what  cost  i  yd.  ? 

24.  If  I  bu.  wheat  cost  14  cts.,  what  cost  -j^q  of  a  bu.  ? 

25.  If  I  yd.  linen   cost  21  cts.,  what  cost  |   of  a  yd.  ? 

Art.  262.  The  solution  of  a  question  by  analysis,  is  an 
analytic  solution.  Every  operation  in  analysis  depends 
directly  on  elementary  or  self-evident  principles. 

To  solve  questions  analytically^  determine  the  process  to  be 
pursued,  by  an  examination  of  the  conditions  of  the  question,  and 
the  relations  which  the  several  quantities  bear  to  each  other. 

In  general,  reason  from  a  given  number  to  unity  (1),  or  one 
part,  and  then  to  the  required  number. 

Art.  263.   PROMISCUOUS   EXAMPLES. 

1.  If  15  bl.  of  flour  cost  $40,  what  cost  24  bl.  ? 

Sol.— If  15  bl.  cost  $40, 1  bl.  will  cost  J^  of  $40=$2|,  and  24  bl. 
■will  cost  24  times  as  much  as  1  bl. ;  24X$2f  =  $64.  Ans. 

Or  thus  :  Since  1  bl.  is  _L  of  15  bl.,  24  bl.  are  ||  of  15  bl.,  and 
will  cost  II  as  much.     $40Xf  |  =  $64.  A7is. 

2.  If  241b.  of  beef  cost  82.16,  what  will  be  the  cost 
of  231b.  ?  Ans.  $2.07 

3.  If  13  yd.  of  cloth  cost  $32.50,  what  will  be  the  cosfc 
ofl4yd.?  Ans.  $35, 

4.  If  $8  will  purchase  4  yd.  of  cloth,  how  many  yards 
will  $24  buy?  Ans.  12yd. 

5.  If  159yd.  of  muslin  cost  $20.67,  how  many  yards 
can  be  bought  for  $34.71  ?  Ans.  267yd. 

6.  If  34  yd.  cost  $147,  what  cost  9  yd.?     Ans  $38||. 

7.  If  5  men  perform  a  piece  of  work  in  12  days,  how 
many  days  will  it  take  3  men  ? 

Solution. — It  will  require  1  man  5  times  as  long  as  5  men ;  that 
is,  5X12  days  =  60  da.  Again,  it  will  require  3  men  1  as  long 
as  1  man  ;  1  of  60  days  =  20  days.  Ans. 

Review. — 2(51.  "What  is  Analysis  ?  262.  On  what  does  every  operation 
in  it  directly  depend  ?    How  ascertain  the  process  to  be  pursued  ? 


264  RAYS    PRACTICAL   ARITHMETIC. 

8.  If  17  men   can  do  a  job  of  work  in  25  days,  in 
what  time  will  10  men  do  it?  Ans.  42^  da. 

9.  If  6  men  can  build  a  wall  in  10  days,  bow  many 
men  can  build  it  in  15  days?  Aiis.  4  men. 

10.  If  5  men  consume  a  barrel  of  flour  in  12  days, 
how  long  would  it  last  4  men  ?  Aris.  15  da. 

11.  A  man  performs  a  piece  of  work  in  7  days,  working 
10  hours  a  day :  if  he  labors  12  hr.  a  day,  how  many 
days  will  be  required?  Ans.  5g  da. 

12.  If  18  men  can  reap  72  acres  of  wheat  in  7  days, 
how  many  days  will  8  men  require?  Ans.  15|  da. 

13.  If  5  hogs  are  worth  9  sheep,  how  many  hogs  will 
pay  for  54  sheep?  Ans.  30. 

14.  If  3  gal.  wine  are  worth  7  gal.  cider,  how  many 
gal.  of  cider  are  worth  42  gal.  wine?  Ans.  98  gal. 

15.  If  a  3  cent  loaf  weigh  8  oz.  when  flour  is  $4  a  bl., 
what  should  it  weigh  when  flour  is  §5  a  bl.?  Ans.  6|  oz. 

16.  If  3  stacks  of  hay  will  feed  12  horses  5  mon.,  how 
long  will  they  feed  20  horses  ?  Ans.  3  mon. 

17.  If  I  of  a  yard  of  gingham  cost  40  cents,  find  the 
cost  of  I  of  a  yard.  Ans.  60  cts. 

18.  If  5  of  a-  yard  of  cloth  cost  $2,  what  will  be  the 
cost  of  I  of  a  yard?  Ans.  ^4.50 

19.  If  I  of  a  tun  of  hay  cost  $4.25,  what  will  |^  of  a 
tun  cost?  Ans.  $3.85 

20.  If  /y  C.  of  wood  cost  $2.52,  what  will  jf  of  a  C. 
cost?  Ans.  $2.86 

21.  If  flb.of  tcacost$.J,what  cost  gib.?     Ans.  U^. 

Examples  in  Art.  2G3  are  often  placed  under  Sim.  Proportion. 
Solve,  now,  by  Analysis,  all  the  questions  in  Art.  203. 

Art.  264.  22.  If  7  men  eat  501b.  of  bread  in  16  days, 
how  many  lb.  will  21  men  eat  in  6  days? 

Solution. — If  7  men  eat  5G  lb.,  1  man  will  eat  ^  as  much, 
•which  is  81b.;  then,  if  1  man  in  10  days  eat  81b.,  in  1  day  he  will 
eat  Jj  of  8  lb.=-p8(j=i  lb.  And,  21  men  will  eat  A-X21=10ilb. 
in  1  day,  and  in  10^XG=G3  lb.  Ans. 


ANALYSIS.  265 

23.  If  2  men  earn  816  in  4  days,  how  much  will  7  men 
earn  in  3  days  ?  Arts.  $42. 

24.  If  2  men  build  12  rods  of  a  wall  in  9  days,  how 
many  rods  can  7  men  build  in  6  days  ?  Ans.  28  rd. 

25.  If  15  oxen  plow  11  A.  in  5  da.,  how  many  oxen 
nil  plow  33  A.  in  9 da.?  Ans.  25  oxen. 

Examples  in  Art.  264  are  Tisually  placed  under  Comp.  Prop. 
Those  in  Art.  205  should  now  be  solved  analytically. 

Art.  265.  26.  A  cistern,  of  250  gal.,  has  two  pipes;  the 
first  yi7/s  41  gal.  an  hour,  and  the  second  empties  6  gal.  an 
hour:  if  both  pipes  are  open,  how  long  will  the  cistern 
be  in  filling?  Ans.  7hr.  8min.  34|sec. 

27.  A  cistern  of  600  gal.  can  be  filled  by  pipe  A,  in 
8  hours;  and  by  pipe  B,  in  12  hours  :  in  what  time  can 
it  be  filled  by  both  pipes?  Ans.  4|hr. 

28.  A  cistern,  of  900  gal.,  can  be  filled  by  pipe  A,  in 
lOhr.,  and  emptied  by  pipe  B,  in  12hr. :  in  what  time 
will  it  be  filled,  if  both  pipes  are  open?  Ans.  60  hr. 

29.  A  can  do  a  job  of  work  in  2  days,  and  B  in  3  days: 
in  what  time  can  both  do  it,  working  together  ? 

Solution. — If  A  does  it  in  2  days,  he  does  ^  of  it  in  1  day: 
if  B  does  it  in  3  days,  he  does  i  of  it  in  1  day;  therefore,  both 
do  2+3=1  ^^  °°^  ^^y»  ^^^  ^^®  whole  in  l-i-|=li  da.  Ans. 

Or,  Suppose  the  work  to  consist  of  any  number  of  equal  parts, 
say  6;  then,  if  A  does  it  in  2  days,  he  will  do  3  parts  in  1  da.;  and 
if  B  does  it  in  3  da.,  he  will  do  2  parts  in  Ida.;  hence,  both 
do  3-}-2=5  parts  in  Ida.,  and  the  whole  in  6-7-5=lida. 

30.  A  can  perform  a  piece  of  work  in  20  da.,  B,  in  15 
da.,  and  C,  in  12da. :  in  what  time  can  the  three  do  it, 
working  together  ?  Ans.  5  da. 

31.  I  hired  4  men  to  build  a  wall :  A  alone  can  do  it 
in  12  da.;  B,  in  15  da. ;  C,  in  18  da. ;  D,  in  24da. :  how 
long  will  it  take  to  do  it,  all  working  ?         Ans.  4:^^  da. 

32.  A  mows  J  of  a  field  in  a  day,  and  B  ^ :  in  what 
time  can  both  do  it  ?  Ans.  4y*^  da. 

33.  A  and  B  can  mow  a  field  in  12  da.,  and  A  alone  in 
20  da.:  in  what  time  can  B  mow  it?  Ans.  30 da. 


266  RAY'S   PRACTICAL  ARITHMETIC. 

34.  A  man  and  his  wife  ate  a  bu.  of  meal  in  6da. ; 
when  the  man  was  absent,  it  lasted  the  woman  15  da. : 
how  long  would  it  last  the  man?  Ans.  10  da. 

35.  A  cistern  has  3  pipes;  the  1st  will  empty  it  in 
4min.,  the  2d  in  8min.,  the  3d  in  15min. :  in  what  time 
will  all  empty  it?  Ans.  2min.  15 i -J  sec. 

36.  A  can  mow  f  of  a  meadow  in  6  days,  and  B,  ^  in 
10  da.:  in  what  time  can  both  mow  it?  Ans.  5|da. 

37.  Divide  35  cents  between  two  boys,  giving  to  one 
9  more  than  the  other.  Ans.  13  and  22. 

Sue. — Subtract  9  from  35,  divide  the  remainder  equally ;  then 
9,  added  to  one  of  the  equal  parts,  will  give  the  greater  share. 

38.  Divide  $3000  into  two  parts,  one  being  $500  more 
than  the  other.  Ans.  S1250  and  §1750. 

39.  A  man  left  $3500  to  his  wife,  son,  and  daughter ; 
to  the  wife,  $800  more  than  to  the  son ;  to  the  son,  $300 
more  than  to  the*  daughter :  find  the  share  of  each. 

Ans.  wife,  $1800;  son,  $1000;  daughter,  $700. 

40.  The  hour  and  minute  hands  of  a  watch  are  together 
at  12  o'clock:  at  what  time  are  they  next  together? 

Sol. — The  min.  hand  moves  over  60  min.  while  the  hour  hand 
moves  over  5  min.;  therefore,  the  min.  hand  moves  over  12  min. 
while  the  hr.  hand  moves  over  1  min.  Hence,  in  moving  over  12 
min.,  the  min.  hand  gains  11  min.  on  the  hr.  hand. 

Now,  12  is  -j-^  of  11,  that  is,  the  distance  moved  over  by  the  min. 
hand,  is  ^^  of  the  distance  gained.  "When  the  min.  hand  is  at  12,  and 
the  hr.  hand  at  1,  the  former  must  gain  5  min.  to  overtake  the 
latter:    -1|  of  5  min.=  |^  =  5j-^j.     A7is.  Dj^'^  min.  past  1. 

41.  At  what  time  between  5  and  6  o'clock  are  the  hr. 
and  min.  hands  together?  Ans.  27-j-^j  min.  after  5. 

42.  At  what  time  between  8  and  9  o'clock,  arc  they 
opposite  to  each  other?  Ans.  10|9  min.  after  8. 

43.  A,  B,  and  C  are  the  partners:  A  put  in  $3276; 
the  j%  of  what  A  put  in  was  equal  to  ^  of  what  B  put  in ; 
and  the  difference  between  A^  of  what  B  put  in,  and 
the  whole  of  what  A  put  in,  equaled  |  of  what  C  put  in. 
They    gained   $7000;    A  received  for   his  share  a  sum, 


ANALYSIS.  2G7 

the  J  of  which  equaled  what  he  put  In ;  C,  a  sum  equal 
to  I  of  what  he  put  in;  and  B  the  remainder.  Find  the 
amount  B  and  C  put  in,  and  each  one's  share  of  the  gain. 

Ans.  B  put  in  $2457;      C,  81820. 
A's  gain  $2730;  B's,  $2086;  C's,  $2184. 

Art.  266.  44.  A  mixes  sugar  at  2cts.  per  lb.,  with 
sugar  at  5  ets.  per  lb.,  so  that  the  mixture  is  worth  3  cts. 
per  lb. :  how  much  of  each  does  he  take  ? 

Solution. — By  taking  1  lb.,  at  2  cts.,  he  gains  1  ct.,  and  by 
taking  lib.  at  5  cts.,  he.  loses  2cts. ;  hence,  that  the  gains  and 
losses  may  be  equal,  he  takes  1  lb.  at  2  cts.,  and  ^  lb.  at  5  cts., 
and  in  the  same  ratio  for  any  quantity  of  the  mixture;  thus,  21b. 
at  2  cts.,  and  lib.  at  Sets.;  41b.  at  2  cts.,  and  21b.  at  o  cts,,  and 
so  on,  will  make  a  mixture  worth  3  cts.  a  lb.     Art.  259. 

45.  In  what  ratio  must  I  mix  sugar  at  4  cts.  a  lb.,  with 
sugar  at  Sets,  a  lb.;  the  mixture  to  be  worth  5  cts.  a  lb.? 

Ans.  31b.  at  4 cts.  to  1  lb.  at  Bets. 

46.  In  what  ratio  must  I  mix  sugar  at  3  cts.  a  lb.,  with 
sugar  at  Sets,  a  lb.;  the  mixture  to  be  worth  Gets,  a  lb.? 

Ans.  21b.  at  3  cts.  to  31b.  at  8 cts. 

47.  How  many  pounds  of  tea,  at  25  cts.  per  lb.,  must 
be  mixed  with  151b.  at  '^0  cts.  per  lb.;  the  mixture  to  be 
worth  28 cts.  per  lb.? 

Solution. — The  ratio  of  the  ingredients  necessary  for  a  mix- 
ture worth  28  cts.  alb.,  shows  that  for  each  1^  lb.  at  30  cts.,  we 
must  take  lib.  at  25 cts.  But  151b. -^1  Jib.  =  10;  hence,  it  will 
take  101b.  at  25  cts. 

48.  How  many  pounds  of  sugar,  at  Sets,  a  lb.,  must  I 
mix  with  101b.  at  11  cts.  a  lb.,  to  make  a  mixture  worth 
10  cts.  alb.?  ^ns.  51b. 

49.  Mix  two  kinds  of  tea  at  20  and  25  cts.  a  lb.,  to 
make  a  mixture  of  251b.  worth  24  cts.  a  lb. 

Solution. — First  find  that  lib.  at  20 cts.  and  41b.  at  25 cts. 
ft  lb.,  make  a  mixture  worth  24  cts.  a  lb.  Then  (Art.  252),  dividing 
25  into  two  parts  having  the  ratio  of  1  to  4,  will  give  5  lb.  at  20cts. 
per  lb.,  and  201b.  at  25  cts.  per  lb. 

The  above  examples  are  usually  placed  under  a  rule  called 
Alligation  Alternate.     They  properly  belong  to  Algebra. 


208  RAY'S   PRACTICAL   ARITHMETIC. 

Art.  267.  50.  The  sum  of  ^,  -J,  and  |  of  a  certain 
number  is  26  :  what  is  that  number  ? 

Sol. — The  sum  of  ^,  1,  and  1  is  1^ ;  hence,  1?  of  the  number  is  26, 
and  Jj  is  J^  of  26  =  2  ;  and  the  number  is  12x2=  24.  Ans. 

51.  One-third  of  a  number  exceeds  ]  of  it  by  8  :  find 
the  number.  Ans.  96. 

52.  Seven-tenths  of  a  number  exceeds  |  of  it  by  7:  what 
is  that  number  ?  Ans.  70. 

53.  I  spent  J  and  |  of  my  money,  and  had  $35  left : 
what  had  I  at  first  ?  Ans.  $75. 

54.  What  number,  increased  by  A,  ^,  and  -|  of  itself, 
gives  the  sum  73  ?  Ans.  30. 

55.  A  boy  lost  J  of  his  money,  spent  20  cts.,  and  had 
15  cts.  left :  how  much  had  he  at  first?  Ans.  63ctb. 

56.  I  spent  |  of  my  money  for  books ;  ?  of  the  rest  for 
paper  ;  I  had  10  cts.  left :  what  had  I  at  first  ?  Ans.  75  cts. 

57.  A  received  $2  for  each  day  he  worked,  and  lost  $1  for 
each  day  idle ;  he  worked  3  times  as  many  days  as  he  was 
idle  ;  at  the  end  of  the  time  he  received  S25  :  how  many 
days  did  he  work?  Ans.  15 da. 

58.  In  an  orchard,  \  the  trees  bear  apples;  ^,  peaches; 
and  ^,  cherries ;  the  remaining  4,  pears :  how  many  trees 
in  the  orchard  ?  Ans.  80. 

59.  A  teacher,  when  asked  the  number  of  his  pupils, 
replied,  that  if  he  had  as  many  more,  half  as  many  more, 
and  1  fourth  as  many  more  as  he  now  had,  he  would 
have  110  :  what  the  number  of  pupils?  Ans.  40. 

60.  A  traveler  spent  the  1st  day,  ^  of  his  money ; 
the  2d  day,  }  of  the  remainder,  and  so  on,  the  3d  and  4th 
days,  when  he  had  $1.62  left:  what  sum  had  he  at  first? 

Ans.  $5.12 

61.  A  merchant  increased  his  capital  the  first  year 
by  T,  of  itself ;  the  2d,  he  increased  this  sum  by  ?  of  itself; 
the  &d,  he  lost  |  of  all  he  had,  which  left  him  $3375.  How 
much  had  he  at  first?  A)is.  $1875. 

Note. — The  examples  in  this  article  are  sometimes  placed  under 
the  rule  of  Position,  now  but  little  used,  as  Algebra  is  generally 
studied  by  the  higher  classes  in  common  schools. 


XXL  EXCHANGE  OF  CURRENCIES. 

Art.  268.  Exchange  or  Reduction  op  Currencies, 
is  the  process  of  changing  the  currency  of  one  country 
to  that  of  another,  without  altering  its  value. 

The  currency  of  a  country  is  its  money  or  circulating  medium. 

Art.  269.  To  change  one  currency  to  another,  the 
different  denominations  in  each  must  be  known ;  also, 
the  unit  value  of  a  denomination  in  one  currency,  in  a 
denomination  of  the  other. 

table   of   ENGLISH,    OR   STERLING   MONEY. 

4  farthings  (far.)  .  make  1  penny, marked  d. 

12  pence 1  shillincr, s. 

20  shillings 1  pound,  orsovereign  ..        £. 

21  shillings 1  guinea,    g. 

Notes. — 1.  Farthings  are  generally  written  as  fractions  of  a 
penny.     Thus,  1  far.  is  written  id.,    2  far.  ^d.,    and  3  far.  |d. 

2.  The  present  legal  value  of  the  pound  sterling,  according  to 
act  of  Congress  of  1842,  is  $4.84 

The  operations  of  Reduction,  Addition,  &c.,  of  sterling  money, 
are  performed  like  those  of  other  Compound  Numbers. 

1.  Reduce  £5  3s.  2id.  to  far.      .     .     Avs.  4953  far. 

2.  Reduce  8675  far.  to  £ Ans.  £9  8|d. 

3.  Add  £3  6 Id.,   £5  10s.  4id.,   £2  15s.  l.}d. 

Ans.  £11  6s. 

4.  £17  6s.  5d.,— £8  5s.  IIH.     .     .  Ans.  £9  S^d. 

5.  Multiply  £3  12s.  2^d.  by  8.  Ans.  £28  17s.  8d. 

6.  £25  10Ad.-r-6,=  what?  Ans.  £4  3s.  5|d. 

7.  Find  the  value  of  £.625  (Art.  188.)     Ans.  12s.  6d. 

Eeview. — 268.  What  is  Exchange  of  currencies  ?  "What  is  the  currency 
of  a  country  ?  2G9.  What  must  be  known  to  change  one  currency  to 
another  ?    Ecpeat  the  table. 

269.  Note.  How  are  farthings  written  ?  What  is  the  legal  value  of 
the  pound  ?  How  are  the  operations  of  reduction,  addition,  &c.,  of  sterling 
money  performed  ? 

269 


270  RAY'S   PRACTICAL  ARITHxMETIC. 

8.  The  value  of  .796875  of  a  £.  Ans.  15s.  ll]d. 

9.  Reduce  7s.  6d.  to  the  decimal  of  a  £.      Ans.  .375 

10.  8s.  9d.  to  the  decimal  of  a£.  (Art.  189.)   Ans.  .4375 

Art.  270.    To  compute  Int.  in  pounds,  shillings,  &c. 

Reduce  the  given  shillings^  pence,  and  farthings,  to  the  deci- 
mal of  a  pound  (Art.  189);  find  the  interest  as  in  dollars  and 
cents  (Art  222);  reduce  the  resulting  decimal  figures  to  shil- 
lings, pence,  and  farthings  (Art.  188). 

11.  What  is  the  interest  of  £75  10s.  for  2  yr.  6  mon., 
at  4  %  ?  Ans.  £7  lis. 

12.  Of  £85  12s.  6d.  for  1  yr.  9  mon.,  at  G  9^  ? 

Ans.  £8  19s.  d^d. 

Art.  271.  1.  Reduce  £12  sterling  to  U.  S.  Money. 

Solution —Since  £1  is  worth  $4.84,  £12  are  worth  12  times  as 
much;  and  §4.84  X  12 =$58. 08  Ans. 

2.  Reduce  £5  6s.  3d.  to  U.  S.  Money. 

Suggestion. —  Either    reduce    the  operation. 

Bhillinjrs.  pence,  and  farthings, to  the  £5  6s.  3d.  =  £5. 3125 

decimul  of  a  pound,   and  multiply  £1  =84.84 

$4.84    by   the    result;    or,   multiply  ^^^^    8*^5   71*^5 

$4.84  by  ihe  pounds,  and  find  the  .     -    . 

value  of  the  lower  denominations,  by  taking  aliquot  parts  (Art.  208). 

3.  Reduce  S40.535  to  sterling  money. 

Solution. — Since  £1  is  $4.84,  there  will  be  as  many  pounds 
in  $40,535  as  $1.84  is  contained  times  in  $40,535 

$40 . 535  -j- $4 . 84  =  8 .  375,  and  £8. 375  =  £8    Ts.  6d.  Ans. 

RULES. 

1.  To  Reduce  Sterling  to  U.  S.  AfoxEY. — Exj^ress  sterling 
money  in  pounds  and  decimals  of  a  pound:  multipli/  this  by  the 
value  of  £\,  ($4.84),  the  product  will  he  the  value  in  dollars. 

Or,  by  Proportion.  As  £1  is  to  the  given  sum,  so 
i.t  $4.84   to  the  value  of  the  given  sum  in  dollars. 

2.  To  n  educe  U.  S.  to  Sterling  ^roxEY — Divide  ihe  given 


EXCHANGE  OF  CURKENCIES.        271 

sum  by  the  value  of£l,  ($4.84),  the  quotient  will  be  the  value 
in  pounds  and  decimals  of  a  pound. 

4.  Reduce  £25  to  U.  S.  money.  Ans.  S121. 

5.  £15  8s.  to  U.  S.  money.  Ans.  S74.536 

6.  £36  15s.  9d.  to  U.  S.  money.  Ans.  $1.78. 05-f 

7.  $179.08  to  sterling  money.  Ans.  £^7. 

8.  $124,388  to  sterling  money.  Ans.  £25  14s. 

9.  In  $1000,  how  many  £?      Ans.  £206  12s.  2|d.-|- 

NoTE. — The  law  of  Congress,  of  July  31st,  1789,  fixed  the  value 
of  the  pound  sterling  at  $4|,  or  $4.44^,  and  $1.  at  48,  6d. 

As  the  then  legal  or  nominal  value  of  the  pound  was  below  its 
real  value,  such  a  per  cent,  was  afterward  added  to  the  nominal  as 
was  necessary  to  make  it  express  the  real  value. 

As  it  requires  nearly  9  per  cent,  of  $4.44|  to  be  added  to  it,  to 
make  $4.84,  when  sterling  funds  or  bills  are  estimated  at  $4^  to  a 
pound,  and  are  9  per  cent,  premium,  they  are  really  only  at  par. 

Art.  272.  In  buying  or  selling  exchange  on  England, 
it  is  still  customary  to  regard  the  pound  as  $4|,  and  then 
to  add  the  ^  premium. 

1.  What  must  be  paid  for  a  bill  of  exchange  on  Lon- 
don of  £200,  at  9  ^  premium  ? 

Solution.— £200  X  4|  =  $888 .  88| ;  $888 .  88f  X  •  09  =  $80  pre- 
mium; $888.88|-t-$80=^$968.88|  Ans. 

2.  What  for  a  bill  of  exchange  on  Liverpool  of  £150, 
at  8  %  premium?  Ans.  $720. 

3.  What  for  a  bill  of  exchange  on  London  of  £80  10s., 
at  n  %  premium?  Ans.  $391.76-]- 

Art.  273.  Previous  to  the  adoption  of  Federal  or 
U.  S.  Money,  in  1786,  accounts  were  kept  in  pounds, 
shillings,  pence,  and  farthings. 

Owing  to  the  fact  that  the  Colonies  had  issued  bills  of  credit^ 

Review. — 270.  How  compute  interest  on  sterling  money  ?  271.  How 
rerluce  sterling  to  U.  S.  Money  ?    How  reduce  IT.  S.  to  sterling  money? 

271.  Note.  What  was  the  legal  value  of  the  pound  sterling  in  1789? 
How  was  the  real  value  found  from  this  ? 


272  RAYS    PRACTICAL    ARITHMETIC. 

which  depreciated  more  or  less,  the  value  of  a  colonial  £  was 
less  than  that  of  a  £  sterling. 

The  depreciation  being  greater  in  some  colonies  than  in 
others,  gave  rise  to  the  different  State  currencies.     Thus, 

In  New  England,  Va.,  Ky.,  and  Tenn.,  Gs.  or  £y3^==$l. 
In  New  York,  Ohio,  and  N.  Carolina,  8s.  or  £  |  =$1. 
In  New  Jersey,  Pa.,  Del.,  and  Md.,  7s.  Cd.  or  £  |  =$1. 
In  South  Carolina  and  Georgia,  48.  8d.  or  £3'''jj— $1. 

In  Canada  and  Nova  Scotia,  5s.         or  £  i  =$1. 

The  process  of  changing  any  sum  of  U.  S.  Money  to 
either  of  these  currencies,  or  the  reverse,  involves  the 
same  principles  as  the  exchanging  of  sterling  money. 

Hence^  to  reduce  U.  S.  Money  to  the  currency  of  a  State, 

Multiply  the  given  sum,  expressed  in  dollars,  hy  the  value 
of  $1  expressed  in  the  fraction  of  a  pound ;  the  product  will 
be  the  value  in  pounds  and  decimals  of  a  pound. 

To  reduce  a  State  currency  to  U.  S.  Money, 

Express  the  given  sum  in  pounds  and  decimals  of  a  pound, 
then  divide  this  hy  the  value  of  $1  expressed  in  the  fraction 
of  a  pound ;  the  quotient  will  he  the  value  in  dollars. 

ANSWERS. 

1.  Reduce  $120.50  to  N.  Eng.  currency.       £36     3s. 

2.  ^75.25  to  N.  York  currency.    .     .     .  £30     2s. 

3.  $98  to  Penn.  currency £36  15s. 

4.  £30  15s.  N.  E.  currency  to  dollars.  .  $102.50 

5.  £25  17s.  N.  Y.  currency  to  dollars.  .  $04,625 

6.  £29    8s.  Georgia  currency  to  dollars.  $126.00 

Note. — Any  sum,  in  one  currency,  may  be  changed  to  that  of 
another,  by  Sim.  Proportion  (Art.  203) ;  or  by  short  methods. 

Thus,  since  G  shillings  New  England  currency  are  equal  to  8  shil- 
lings New  York  currency,  to  change  the  former  to  the  latter,  add 
one-third  of  the  sum;  to  change  the  latter  to  the  former,  subtract 
one-fourth  of  the  sum. 

Review. — 272.  In  buying  or  selling  bills  of  exchange,  how  make  the 
calculations?     273.  In  what  wore  accounts  kept  previous  to  178'j  ? 


DUODECIMALS.  273 

Art.  274.  Any  currency  may  be  reduced  to  U.  S.  Money, 
or  U.  S.  Money  to  any  currency,  by  multiplication  or  divi- 
sion, as  in  Art.  271,  when  the  value  of  a  unit  of  the  foreign 
currency  expressed  in  U.  S.  Money  is  known. 

The  unit  of  French  money  is  the  franc,  its  value  being  $0,186 
Bills  of  exchange  on  France  are  bought  and  sold  at  a  certain 
number  of  francs  to  the  dollar. 

EXAMPLE. — At  $1  for  5.30  francs,  what  will  be  the 
cost  of  a  Bill  on  Paris  for  1166  francs?  Ans.  8220. 

In  ^^Raifa  Higher  Arithmetic "  may  be  found  a  complete  table  of  all 
foreign  coins,  with  their  value  in  U.  S.  Money  ;  also,  valuable  infonnation 
respecting  exchange  with  all  civilized  nations. 


XXII.    DUODECIMALS. 

Art.  275.  Duodecimals  are  a  peculiar  order  of  frac- 
tions, which  increase  and  decrease  in  a  twelve-fold  ratio. 

Their  name,  from  the  Latin  duodecim,  signifies  twelve. 

The  unit,  1  foot,  is  divided  into  12  equal  parts,  called  inches, 
or  primes,  marked  thus,  (^). 

Each  inch,  or  prime,  is  divided  into  12  equal  parts,  called 
seconds^  marked  (^^). 

Each  second,  into  12  equal  parts,  called  thirds,  {^^^'y 

Each  third,  into  12  equal  parts,  called /bwr^As,  (^^^0- 

Hence,  V    inch,  or  prime,     .     .     .     =    yV     of  a  foot. 

y^  second  is  yly  of  J.T    .     .     =  jl^    of  a  foot. 

1--  third  is  -^  ;f  Jj  ;f  J^  .     =  -^ijg  of  a  foot. 

TABLE. 

12  fourths  (•'^^^)    ....  make  1  third,    .....  marked  ^^^. 

12  thirds     1  second, ''''. 

12  seconds 1  inch  or  prime,       ..  ''. 

12  inches  or  primes   .       ..     1  foot, ft 

The  marks  ^,  ^^,  ^^■',  '''''''',  called  indices,  show  the  different  parts. 
3d  Bk.  18 


274  RAY'S   PRACTICAL   ARITHMETIC. 

Duodecimals  are  added  and  subtracted  like  Compound  Num- 
bers; 12  units  of  each  order  making  a  unit  of  the  next  higher. 

Art.  276.   multiplication. 

Duodecimals  are  used  in  the  measurement  of  surfaces 
and  solids^  as  boards,  solid  walls,  &c. 

1.  Find  the  superficial  contents  of  a  board  7ft.  Sin. 
long,  and  4ft.  Sin.  wide. 

Length,  multiplied  by  breadth,  gives  the  superficial  contents. 

Solution. — 5''=  -Af  of  a  foot ;  therefore, 
5^X4  ft.  =  5x4=^20  =  20  inches,  which 
is  1  ft.  8  in. ;  write  the  inches  or  primes  in  the 
order  of  primes,  and  carry  the  1  ft. 

Next,  7  ft,  X  4  ft.  =  28  ft.,  to  which  add 
the  1  ft.  carried,  the  sum  is  29  ft. ;  which  write 
in  the  order  of  feet. 

Again,    5^=  -^^,    and    8''=  -''-,-  ;    therefore, 

B^X^'=j%Xj\  =  jW^l'y,  which  is  V    3^^; 
in  the  order  of  seconds,  and  carry  the  1  in. 

Next,  7  ft.  X  3  in.  =  7  X  t^^  =  H  =  21^  and  V  carried, 
make  22^=:  1  ft.  10^  Writing  these  in  their  orders,  and  adding 
the  two  products,  the  entire  product  is  31  ft.  6^  3^^. 


OPERATION. 

ft. 

/ 

7 

5 

4 

3 

29 

8 

1 

10  3" 

Ans.  31 

6  3 

;  write  the  3   sec. 

The  pre  duct  of  aiuj  two  denominations,  is  of  that  denomina- 
tion denoted  by  the  sum  of  their  indices  ;  thus,  3^X  ^^  =  1^^^ 
3'X7-^2r,   7-X4^  =  -p|5Xr1z  =  Tfi5  =  28---    Hence, 

feet        multiplied  by  feet,         give  (square)  feet 
Feet         multiplied  by  inches,     give  inches. 
Inches     multiplied  by  inches,     give  seconds. 
Inches     multiplied  by  seconds,  give  thirds. 
Seconds  multiplied  by  seconds,  give  fourths,  and  so  on. 

Review. — 273.  Why  was  tho  value  of  a  colonial  pound  less  than  that 
of  a  pound  sterling?    How  reduce  U.  S.  Money  to  a  State  currency  ? 

27.3.  IIow  reduce  a  State  currency  to  U.  S.  Money  ?  275.  What  arc 
duodecimals?  Whence  their  name?  What  are  Primes?  Sccoads? 
Thirds?  Fourths?  Repeat  the  table.  What  arc  indices?  What  part  of 
afoot  ial'?  1"?  1'"?  1""?    IIow  are  duodecimals  added  and  subtracted  ? 


INVOLUTION.  275 

Rnle  for  Multiplication. — 1.  Write  the  multiplier  under  the 
multiplicand^  placing  units  of  the  same  order  under  each  other. 

2.  Multiply,  first  hy  the  feet,  next  by  the  inches,  and  so  on, 
recollecting  that  the  product  will  be  of  that  denomination  denoted 
hy  the  sum  of  their  indices. 

3.  Add  the  several  partial  products  together,  and  their  sum 
will  be  the  required  product. 

The  primes  of  the  product  of  two  duodecimal  factors,  are  neither 
linear  nor  square  in.,  but  twelfths  of  a  sq.  ft.  The  primes  of  the  pro- 
duct of  three  duodecimal  factors,  are  twelfths  of  a  cu.  ft. 

2.  How  many  square  feet  in  a  board  5ft.  Sin.  lonsr, 
and  1  ft.  5  in.  wide  ?  Ans.  7  sq.  ft.  5'  3".  " 

3.  Multiply  5 ft.  7 in.  by  1ft.  10 in.  ^/is.  10 sq.ft.  2' 10^ 

4.  8  ft.  6  in.  9"  X  7  ft.  3  in.         Ans.  62  sq.  ft.  11'^  3'". 

5.  8ft.  4 in.  6"X2ft.  7 in.  4''.     Ans.  21  sq.ft.  10'  5''. 

6.  4ft.  5'  6"X2ft.  3'  5".  Ans.  lOsq.ft.  2'  2"  9'"  6' 

Another  method  of  solution  found  in  "Eay^a  Higher  Arithmetic.' 


}ttH 


XXIII.    INVOLUTION. 

Art.  277.  Involution  is  the  multiplication  of  a  num- 
ber by  itself  one  or  more  times. 

A  POWER  is  the  product  obtained  by  involution. 

The  ROOT,  or  first  power ^  is  the  number  multiplied. 

If  the  number  be  taken  twice  as  a  factor,  the  product  is  the 
second poiver ;     3X3  =  9,  is  the  2d  power  of  3. 

If  the  number  be  taken  3  times  as  a  factor,  the  product  is 
the  3d  power ;     2X2X2  =  8,  is  the  3d  power  of  2. 

Eeview. — 276.  For  what  are  duodecimals  used  ?  Of  what  denomina- 
tion is  the  product  of  any  two  denominations  ?  What  is  tho  product  of 
feet  by  feet?  Feet  by  inches?  Inches  by  inches?  Inches  by  seconds? 
Seconds  by  seconds  ?  Eulc  for  multiplication  ?  What  do  tho  primes  of 
the  product  of  two  duodecimal  factors  represent  ?    Of  three  ? 


276  RAY'S    PRACTICAL    ARITHMETIC. 

And,  if  taken  4  times  as  a  factor,  the  product  is  the  4th 
power;  if  5  times,  the  product  is  the  5th  power,  and  so  on. 

Hence,    the   different  powers    derive  their  name  from  the 
number  of  times  the  root  is  taken  as  a  factor. 

Rem. — The  given  number  is  called  the  root^  the  diflFerent  powers 
of  the  Humber  being  derived  from  it. 


'o 


Art.  278.  The  second  power  of  a  number  is  called 
the  square;  the  third  power  the  cube.  These  terms  are 
derived  thus: 

Illustrations.  1.  Take  a  line,  say  3  feet  long,  its  Jirst  power 
is  the  line  itself 

2.  If  3  feet  be  multiplied  by  itself,  the  product  (Art.  87)  will 
be  3  X  3=9  square  feet.  (See  diagram  of  3  feet  square,  page  91.) 
But  3X  3  =  9  is  the  second  power  of  3;  hence,  the  2d  power  is 
called  the  square. 

3.  If  each  side  of  a  cube  is  3  feet,  the  cube  (Art.  92)  contains 
3X  3  X3  =  27cMMcft.  (Seediagram,  p.  94.)  But 3X  3  X  3  =  27,  is 
the  third  power  of  3 ;  hence,  the  3d  power  is  called  the  cube. 

Art.  279.  The  number  denoting  the  power  to  which 
the  root  is  to  be  raised,  is  the  index  or  exponent  of  the 
power.  It  is  placed  on  the  right,  a  little  higher  than  the 
root.     Thus, 

2^  =2,  the  1st  power  of  2. 
2^  =  2X2=^ 4,  the  2d  power  or  square  of  2. 
2  3=2X2X2  =  8,  the  3d  power,  or  cube  of  2. 
2  4=2X2X2X2=^16,  the  4th  power  of  2,  &c. 

To  find  the  second  power  of  2,  use  it  as  a  factor  twice ; 
thus,  2X2  =  4.  To  find  the  third  power  of  2,  use  it  three 
times;  thus,  2X2X2  =  8,  and  so  on. 


Review.— 277.  What  is  involution  ?  "What  is  a  power  ?  "What  is  the 
root,  or  1st  power  ?    The  2d  power  ?    The  8d  ?    The  4th  ? 

277.  From  what  do  the  different  powers  derive  their  name  ?  Rem.  "Why 
is  the  given  number  called  the  root  ?  278.  "What  is  the  second  power  of  a 
number  called  ?  "What  the  3d  ?  How  are  these  terms  derived  ?  279.  "What 
is  the  index  of  a  power?  How  find  the  2d  power  of  2  ?  The  3d?  The  4th? 


EVOLUTION.  277 

Rule  for  Involution. — Multiply  the  number  by  itself,  till 
it  is  used  as  a  factor  as  many  times  as  there  are  units  in  the 
index  of  the  power. 

EXAMPLES    FOR    MENTAL   SOLUTION. 

1.  What  is  the  square  of  1?     of    2?     of    3?   of    4? 

of  5?  of  6?  of  7?  of  8?  of  9?  of  10?  of  11? 

2.  What  is  the  cube      of  1?  of  2?  of    3?  of    4? 

3.  What  is  the  square  of  ;V?  of  J  ?  of    §?  of    ^? 

4.  What  is  the  cube      of  i  ?  of  ^  ?  of    §  ?  of    |  ? 

5.  What  is  the  fourth  power  of  2  ?  fifth  power  of    2  ? 
fourth  power  of  3? 

SLATE    EXEHCISES. 

6.  What  is  the  2d  power  or  sq.  of  65?  Ans.    4225. 

7.  The  third  power  or  cube  of  25?       Ans.  15625. 

8.  The  square  of  16|? .     .     .     .     .     Ans.    272^. 

9.  The  cube  of  121  ?'^ ^«s.  1953^. 

10.  The  fourth  power  of  13?    .     .     .     ^/zs.  28561. 

11.  The  fifth  power  of  |?    .     .     .     .         Ans.  ^^\. 

12.  The  sixth  power  of  9?  .     .     .     .  ^7is.  531441. 

13.  The  4th  power  of  .025?     Ans.  .000000390625 

14.  The  value  of   143? jins.      2744. 

15.  The  value  of   194? ^ns.  130321. 

16.  The  value  of   2J 5? Ans.    69^\%. 

17.  The  value  of  .09^?  .     .      Ans.  .000000531441 


XXIV.    EVOLUTION. 

Art.  280.  Evolution,  or  the  extraction  of  roots,  is 
the  process  of  resolving  numbers  into  equal  factors. 

When  a  number  is  resolved  into  equal  factors,  each  factor 
is  a  ROOT  of  the  number. 

Hence,  a  root  is  a  factor  which,  multiplied  by  itself  a  certain 
number  of  iimeSf  will  produce  the  given  number. 


278  RAY'S   PRACTICAL  ARITHMETIC. 

One  of  the  two  equal  factors  of  a  number,  is  the  second  root, 
or  square  root  of  that  number.     Thus, 

9  =^  3  X  3  ;  three  being  the  square  root  of  9. 
One  of  the  three  equal  factors  of  a  number,  is  the  third,  or 
cube  root  of  that  number.     Thus, 

8  =  2X2  X  2;  two  being  the  cube  root  of  8. 

Also,  one  of  the  foicr,  Jive,  &c.,  equal  factors  of  a  number  is 
the  fourth,  Jifth,  kc,  root  of  that  number. 

Hence,  the  name  of  the  root  shows  the  number  of  equal 
Sactors  into  which  the  given  number  is  resolved.     Thus. 

The  square  root  of  25  is  5,  as  5  X  5  =  25. 

The  cube     root  of  27  is  3,  as  3  X  3  X  3  =  27. 

"YhQ  fourth  root  of  16  is  2,  as  2  X  2  X  2  X  2  =  16,  &c. 

Evolution  is  the  reverse  of  Involution.  In  Involution,  the  root  is 
given  to  find  the  power ;  in  Evolution,  the  power,  to  find  the  root. 

When  one  number  is  Vk  power  of  another,  the  latter  is  a  root  of 
the  former:  thus,  8  is  the  cube  of  2,  and  2  is  the  cube  root  of  8. 

Art.  281.  Roots  are  Denoted  in  Two  Ways  : 

1st.  By  |/  called  the  radical  sign,  placed  before  the  number. 
2d.    By  a  fractional  index  placed  on  the  right  of  the  number. 

Thus,  \/  4,  or  4^,  denotes  the  square  root  of  4. 

And,  \l  21,  or  27^,  the  cube  root  of  27. 

Note. — The  figure  over  the  radical  sign,  denotes  the  name  of 
the  root.  When  the  sign  has  no  figure  over  it,  2  is  understood ; 
thus,  'i/25  and  v^25,  each  denotes  the  square  root  of  25.  The 
denominator  of  the  fractional  index  denotes  the  name  of  the  root. 

Art.  282.  Any  number  whose  exact  root  can  be 
obtained,  is  a  perfect  power:  as,  4,  9,  16,  &c. 

Review.— 279.  What  ia  the  rule  for  involution?  280.  What  is 
evolution?  What  is  a  root  of  a  numbf^r?  What  t ho  second  or  square 
root?  What  the  third,  or  cube  root?  What  does  the  name  of  the  root 
show  ?  Give  examples.    Rem.  Why  is  evolution  the  reverse  of  involution  1 

281.  IIuw  are  roots  denoted?  Give  examples.  Note.  What  does  the 
figure  over  the  radical  sign  denote?  What  the  denominator  of  the  frac- 
tional index?     282.  What  is  a  perfect  power  ?    Give  examples. 


SQUARE    ROOT.  279 

Art.  283.  Every  root,  and  every  power  of  1,  is  1. 
Thus,  i/l,  l/i,  Vl>  and  (1)2,  (1)3,  (1)4^  each=-l. 

EXTRACTION  OF  THE  SQUARE  ROOT. 

Art.  284.  To  extract  tlie  square  root  of  a  number,  is 
to  resolve  it  into  two  equal  factors  (Art.  280)  ;  or,  to 
find  a  number,  which,  multiplied  by  itself,  will  produce 
the  given  number.  4  j^_  ^^^^^ 

The  extraction  of  the  square  root  (Art.  278)  I                      I  q^. 

is  also  the   method  of  finding  the  number  of p* 

units  in  the  side  of  a  square,  when  its  super- ^ 

ficial   contents   are    known;    or,    knowing  the pi 

superficial  contents,   it  shows  how  to  arrange       

them  so  as  to  form  the  largest  square  possible.  4  X4  =  16. 

EXAMPLE. — What  is  the  side  of  a  square  board  which 
contains  16  square  inches? 

Solution. — Since  16  =  4  X  4,  each  side  is  4  inches. 

Art.  285.  The  first  ten  Numbers  and  their  Squares  are 

Numbers.  1,  2,  3,  4,  5,  6,  7,  8,  9,  10. 
Squares.  1,  4,  9,  16,  25,  36,  49,  64,  81,  100. 
The  numbers  in  the  1st  line  are  the  square  roots  of  those  in  the  2d. 

Since  the  sq.  root  of  1  is  1,  and  of  100  is  10,  the  sq. 
root  of  any  number  less  than  100  consists  of  one  figure. 

That  is,  the  square  root  of  a  number  of  fewer  than  three 
figures,  must  consist  of  only  one  figure. 

Again,  take  the  numbers  10,  20,  30,  40,  &c.,  to  100. 

Their  squares  are  100,  400,  900,  1600,  &c.,  to  10000. 

Since  the  square  root  of  100  is  10,  and  of  10000  is 
100,  the  square  root  of  any  number  greater  than  100, 
and  less  than  10000,  will  consist  of  tico  figures. 


Eev.— 283.  What  are  the  different  roots  and  powers  of  1  ?  284.  What 
is  it  to  extract  the  square  root  of  a  number  ?  What  else  may  it  be  con- 
sidered ?  What  is  the  side  of  a  square  containing  9sq.  in.?  25 sq.  in.? 
285.  What  the  rule  for  pointing  ?    Why  ? 


280 


RAY'S    PRACTICAL   ARITHMETIC. 


256(10-1-6 
100  =16 
Ans. 


10156 
2156 


The  square  root  of  a  number  of  more  than  two  figures  and 
fewer  than^ye,  must  consist  of  two  figures. 

Also,  the  square  root  of  a  number  of  more  than  four 
figures  and  less  than  seven,  must  consist  of  three  figures. 

Hence,  if  a  dot  (.)  be  placed  over  every  alternate  figure, 
beginning  with  units,  the  number  of  dots  will  be  the  number 
of  figures  in  the  root.     This  is  the  rule  for  pointing. 

Art.  286.  1.  Extract  the  square  root  of  256  ;  or,  wliat 
is  the  same,  arrange  256sq.in.  in  the  form  of  a  square. 

Sol. — To  ascertain  the  number  of  figures  operation. 

in  the  root,  begin  at  the  unit's  place,  and 
place  a  dot  over  each  alternate  figure.  This 
shows  that  the  root  consists  of  two  figures. 

Next  find  that  the  largest  square  in  2  (hun- 
dred) is  1  (hundred),  the  sq.  root  of  which        20 
is  1  (ten),  which  putontheright,  as  in  writing  6 

Ihequotientindivision.  Subtract  the  100  from        26 
the  given  number,  and  156  remain.     While 
Bolving  this  example  by  figures,  attend  to  arranging  the  squares. 

After  finding  that  the  sq. 
root  of  the  given  number 
will  contain  two  places  of 
figures,  (tens  and  units) 
and  that  the  figure  in  tens' 
place  is  1  (ten) ;  form  a 
square  figure,  A,  10  in.  on 
each  side,  which  contains 
(Art.  87)  lOOsq.in. ;  taking 
this  sum  from  the  whole 
number  of  squares,  156  sq. 
in.  remain,  which  correspond 
to  the  number,  156,  left  after 
subtracting  above. 


J3 

JD 

•^ 

<o 

:  io)((^=6o 

6/6=36 

^ 

-      A. 

C 

>«^ 

^ 

•5 

-  10  )( 70=100 

70/6=60 

"^> 

"^ 

v^: 

_ 

1    /   f  r   f   f   f   /  f 

1   r  r  1  1 

10  ific/ics 


d  Cn 


Without  the  diagram,  it  would  be  difficult  to  tell  what  operation 
to  perform  on  the  156  to  obtain  the  other  figure  of  the  root. 

By  examining  the  figure  A,  it  is  obvious,  that  to  increase  it,  and 
at  the  same  time  preserve  it  a  square,  both  length  and  breadth 
must  be  increased  equalli/ ;  and  since  each  side  is  lOin.  long,  it  will 


B 

G 

D 

SQUARE  ROOT.  281 

take  twice  10,  that  is,  20  in.  to  encompass  two  sides  of  the  square  A. 
For  this  reason,  10  is  doubled  in  the  numerical  operation. 

Now  determine  the  breadth  of  the  addition  to  be  made  to  each 
side  of  the  square  A. 

By  examining  the  figure,  we  see  that  after  increasing  each  side 
equally,  it  will  require  a  small  square,  D,  of  the  same  breadth  as 
each  of  the  figures  B  and  C,  to  complete  the  entire  square. 

Hence,  the  superficial  contents  of  B,  C,  and  D,  must  be  equal  to 
the  remainder  (156) ;  now  the  contents 
of  B,  C,  and  D,  are  obtained  by  mul- 
tiplying their  whole  length  by  the 
breadth  of  D;  this  is  made  clearer  by 
examining  the  figure  in  the  margin. 

It  is  now  obvious  that  the  figure  (6)  in  the  units'  place,  that  is, 
the  breadth  of  B  and  C,  must  be  found  by  trial,  and  that  it  will  be 
somewhat  less  than  the  number  of  times  the  length  of  B  and  C  (20) 
is  contained  in  the  remainder  (156).  20  is  contained  in  156  more 
than  7  times ;  let  us  try  7 : — 7  added  to  20  makes  27  for  the  whole 
length  of  B,  C,  and  D,  and  this,  multiplied  by  7,  (Art.  88),  gives 
189  for  the  superficial  contents  of  B,  C,  and  D ;  which  being  more 
than  156,  the  breadth  (7)  was  taken  too  great. 

Next,  take  6  for  the  length  and  breadth  of  D,  and  adding  this 
to  20,  gives  26  for  the  length  of  B,  C,  and  D ;  multiplying  this  by 
the  breadth  (6)  gives  156  for  the  superficial  contents  of  B,  C,  and  D. 

Hence,  the  square  root  of  256  is  16 ;  or,  when  256  sq.  in.  are 
arranged  in  the  form  of  a  square,  each  side  is  16  inches. 

In  performing  the  operation  numerically  operation. 

as  in  the  margin,  it  is  not  necessary  to  256(16=root, 

place  a  cipher  on  the  right  of  the  first  1 

figure  of  the  root,  either  before  or  after  

doubling  it,  since  that  place  is  to  be  filled     ^  t>  j  i  O  o 
by  the  next  figure  of  the  root.  1 0  o 

A  contains  1  0  X  1  0=1  0  0  sq.  in.  T^e  superficial  contents 

xJ         ..         10Xd=60      ..  of  the  several  parts  of  the 

O         ..         lOX  .6:=    6  0      ..  figure  added  together,  are 

■U         ,.            oX     6=    oD      ..  equal  to  those  of  the  whole 

1  py-i  p ^^fi  found  by  squaring  one  side. 

Another  explanation  of  this  subject  is  in  '^Ray's  Higher  Arithmetic.''* 


282  RAY'S   PRACTICAL    ARITHMETIC. 

*2.  Find  the  square  root  of  529.  Ans.  23. 

3.  Find  the  square  root  of  5664-4.  operation. 

Sol.— By  the  Rule  for  Pointing,  (Art.  285),             56644(238^ 
there  are  three  periods;  hence,  the  root  will             4 
consist  of  three  places  of  figures.  

In  performing  the  operation,  1st  find  the  'i^jioo 
sq.   root  of  566  as  in  example    2d.     Next  i.  ^J 

consider  23  as  so  many  tens,  and  find  the  468')3744 
last  figure  (8)  as  the  figure  3  was  found.  3744 

4.  The  sq.  root  of  915.0625  operation. 

Sol. — The  product  of  two  decimals              915   0625^30.25 
will  contain  as  many  decimal  places               n 
as  there  are  decimals  in  both  factors • 

(Art.  187) ;  if  the  root  has  one  deci-      602)1506 

1  '^  0  4 

mal  place,  its  square  will  have  two ;  ^  -'  ^'* 

if  the  root  has  two  decimal  places,  its      6045^30225 
square  will  have  four.  30225 

Hence,  in  pointing  a  decimal  number  to  extract  its  square  root,  put 
a  dot  over  each  alternate  place,  beginning  with  tenths. 

In  solving  this  example,  the  6  (tens)  are  not  contained  in  15; 
therefore  place  a  cipher  in  the  root  and  bring  down  another 
period.  Hence,  the  square  root  of  decimals  is  extracted  in  the 
same  manner  as  that  of  whole  numbers. 

5.  What  is  the  square  root  of  |? 
Solution. — Since  |  =  |  X  |i  the  square  root  of  ^  is  |. 

Hence,  the  square  root  of  a  common  fraction  is  equal  to  the  square 
root  of  its  numerator  divided  by  the  square  root  of  its  denominator. 

Art.  287.  TO  extract  the  square  root, 

Kule. — 1.  Separate  the  given  number  into  periods  of  two 
places  each,  by  placing  a  dot  over  the  alternate  figures,  beginning 
with  units.  (The  left  period  will  often  have  but  one  figure.) 

Review. — 286.  How  find  the  first  figure  of  the  root  ?  Why  double  the 
first  figure  for  a  trial  divisor  ?    IIow  find  the  second  figure  of  the  root  ? 

286.  "VY^hy  aild  this  to  the  trial  divisor  to  get  the  complete  divisor  ?  How 
point  a  decimal  number  to  extract  its  square  root  ?    Why  ? 


SQUARE   ROOT. 


283 


2.  Find  the  greatest  square  in  the  left  period,  and  place  its 
root  on  the  right,  like  a  quotient  in  division.  Subtract  the 
square  of  this  root  from  the  left  period,  and  to  the  remainder 
bring  doicn  the  next  period  for  a  dividend. 

3.  DoiLble  the  root  found,  and  place  it  on  the  left  for  a  trial 
divisor.  Find  how  many  times  the  divisor  is  contained  in  the 
dividend,  exclusive  of  the  right  hand  figure,  and  place  the  quo- 
tient in  the  root  and  also  on  the  right  of  the  divisor. 

4.  Multiply  the  divisor  thus  increased,  by  the  lastfgure  of  the 
root;  subtract  the  product  from  the  dividend ;  to  the  remainder 
annex  the  next  period  for  a  new  dividend. 

5.  Double  the  whole  root  found  for  a  new  trial  divisor,  and 
continue  the  operation  as  before,  until  all  the  periods  are  used. 

Notes. — 1.  If  any  trial  divisor  is  not  contained  in  its  dividend, 
place  a  cipher  in  the  quotient,  and  bring  down  another  period. 

2.  After  bringing  down  the  last  period,  and  finding  the  figure  of 
the  root  corresponding  to  it,  if  there  is  a  remainder,  periods  of 
ciphers  may  be  annexed,  and  the  operation  continued  at  pleasure. 

3.  To  extract  the  sq.  root  of  a  common  fraction  reduce  it  to  its 
lowest  terms  (Art.  138),  then  extract  the  sq.  root  of  the  numerator 
and  denominator.  If  both  terms  are  not  perfect  squares,  reduce 
it  to  a  decimal,  then  extract  the  sq.  root. 

To  extract  the  sq.  root  of  a  mixed  number,  first  reduce  it  to  an 
improper  fraction;  or,  reduce  the  fractional  part  to  a  decimal. 


What  is  the  Square  Root  of, 


6.  625? 

7.  6561? 

8.  390625? 

9.  1679616? 

10.  5764801? 

11.  43046721? 

12.  987656329? 

13.  289442169? 

14.  234.09? 

15.  145.2025? 


Ans.  25.  16. 

81. 1 17. 

625. :18. 

1296.  j 19. 

2401. '20. 

6561. I  21. 

31427.  22. 

17013.  23. 

15.3  24. 

12.05  25. 


.0196?    Ans.   .14 

1.008016?     1.004 

.00822649?   .0907 


2_5  ? 

_8J  7_  ? 
1  183  * 

30|? 
10? 

2? 

.2? 
3  • 

6^? 


5 

1  1 
T3' 

3.162277-f- 

1.41421-1- 

.81649-1- 

2.5298-f- 


284 


RAY'S  PRACTICAL   ARITHMETIC. 


26,  How  many  rods  on  each  side  of  a  square  field 
of  6241  sq.  rd.?  Ans.  79  rds. 

THE    SQUARE    ROOT    BY    FACTORING. 

Art.  288.  Since  any  square  is  the  product  of  two  equal 
factors,  if  a  perfect  square  be  separated  into  its  prime  fac- 
tors (Art.  113),  its  square  root  will  be  composed  of  half 
the  equal  factors.     Thus, 

441=3X3X7X7;  hence  ^441=3x7=21. 

Therefore^  to  obtain  the  square  root  of  any  perfect  square, 
resolve  the  number  into  its  prime  factors,  and  take  the  product 
of  one  of  each  two  equal  factors. 

By  Factoring,  find  the  Square  Root 


1. 

Of  16. 

Ans.    4. 

5. 

Of  16X25. 

Ans.  20. 

2. 

36. 

Ans.    6. 

6. 

36X49. 

Am.  42. 

3. 

100. 

Ans.  10. 

7. 

64x81. 

Ans.  72. 

4. 

225. 

Ans.  15. 

8. 

121X25. 

Ans.  55. 

Art.  289.  applications  of  the  square  root. 

DEFINITIONS. — A  triangle  is  a  figure  bounded  by  three 
straight  lines.  When  one  side  is  perpendicular  to  another, 
the  angle  between  them  is  a  right  angk,  and  the  triangle 
is  a  right-angled  triangle. 

The  side  opposite  the  right  angle  is  called 
the  hypotenuse;  the  other  two  sides,  the 
base  and  perpendicular. 

Thus,  A  B  C  is  a  right-angled  triangle; 
A  B  being  the  base,  B  0  the  perpendicular, 
and  A  C  the  hypotenuse. 

Art.  290.  It  is  a  known  principle,  that  the  square  of 
the  hypotenuse  of  a  right-angled,  triangle  is  equal  to  the  sum 
of  the  squares  of  the  other  two  sides. 

Review. — 28Y.  "What  is  tho  rule  for  square  root?  Notes.  How  proceed 
when  any  trial  divisor  is  not  contained  in  tho  corresponding  dividend? 

287.  When  there  is  a  remainder  after  bringing  down  tho  hist  period,  how 
proceed  ?  IIow  extract  tho  square  root  of  a  common  fraction  ?  IIow,  if 
both  terms  are  not  perfect  squares  ?    IIow,  if  a  mixed  number  ? 


SQUARE  ROOT. 


285 


^yv- 

^/ 

y 

^ 

B 

Thus,  in  the  triangle  ABC,  the  side  A  B 
is  3  feet;  B  C,  4  feet ; "and  A  C,  5  feet;  the  sum 
of  the  squares  of  A  B  and  B  C,  9+16,  is  equal 
to  25,  the  square  of  the  side  AC.  a 

Rule    for     finding    the    hypotenuse 
when   the  base  and  perpendicular 
are  known. 

To  the  square  of  the  hase^  add  the 
square  of  the  perpendicular ;  the  square 
root  of  the  sum  will  give  the  hypotenuse. 

Rule  for  finding  either  side  when  the  hypotenuse  and 
the  other  side  are  known. 

From  the  square  of  the  hypotenuse,  subtract  the  square  of  the 
other  given  side;  the  square  root  of  the  remainder  will  give  the 
required  side. 

1.  The  base  and  perpendicular  of  a  right-angled  tri- 
angle are  30  and  40:  what  the  hypotenuse?       Ans.  50. 

2.  The  base  and  perpendicular  of  a  right-angled  tri- 
angle are  81  and  108:  what  the  hypotenuse?   Ans.  135. 

3.  The  hypotenuse  of  a  right-angled  triangle  is  100 ; 
the  base,  60:  what  the  perpendicular?  Ans.  80. 

4.  A  castle  45  yd.  high,  is  surrounded  by  a  ditch  60  yd. 
wide :  what  length  of  rope  will  reach  from  the  outside  of 
the  ditch  to  the  top  of  the  castle  ?  Ans.  75  yd. 

5.  It  is  36  yd.  from  the  top  of  a  fort,  standing  by  the 
edge  of  the  water,  to  the  opposite  side  of  a  stream  24  yd. 
wide:  how  high  is  the  fort?  Ans.  26.83-}- yd. 

6.  A  ladder  60  ft.  long,  reaches  a  window  37  ft.  from 
the  ground  on  one  side  of  the  street,  and  without  moving 
it  at  the  foot,  will  reach  one  23  ft.  high  on  the  other  side ; 
find  the  width  of  the  street.  Ans.  102.644-  ft. 

Review. — 28S.  How  extract  tho  square  root  of  a  perfect  square  by 
factoring  ?  289.  "What  is  a  triangle  ?  A  right  angle  ?  A  right-angled 
triangle  ?     "What  the  hypotenuse  ?     "What  the  other  two  sides  ? 

290.  "What  is  true  of  every  right-angled  triangle  ?  Give  an  example. 
How  find  the  hypotenuse  when  the  base  and  perpendicular  are  known  ? 
How  find  either  side  when  the  hyiwtenuse  and  the  other  side  are  known  ? 


286  RAY'S    PRACTICAL    ARITHMETIC. 

7.  A  tree,  140  ft.  high,  is  in  the  center  of  a  circular 
ishmd  100ft.  in  diameter;  a  line  600  feet  long,  reaches 
from  the  top  of  the  tree  to  the  further  shore  :  what  is  the 
breadth  of  the  stream,  the  land  on  each  side  being  of  the 
same  level?  ^ws.  533. 434- ft. 

8.  A  room  is  20ft.  lonir,  16ft.  wide,  and  12ft.  his;!! : 
what  is  the  distance  from  one  of  the  lower  corners  to  the 
opposite  upper  corner?  Ans.  28. 28-}- ft. 

Art.  291.  Since  the  area  or  superficial  contents  of  a 
square  equals  the  square  of  one  of  its  sides,  (Art.  87), 
hence,  the 

Rule  for  finding  the  side  op  a  squaee  equal  in  area 
to  any  given  surface. 

Extract  tJie  square  root  of  the  given  area;  the  root  will  be 
the  side  of  the  required  square. 

1.  The  superficial  contents  of  a  circle  are  4096  :  what 
the  side  of  a  square  of  equal  area?  Ans.  64. 

2.  A  square  field  measures  4rd.  on  each  side:  what 
the  length  of  one  side  of  a  square  field  having  9  times  as 
many  sq.  rd.  ?  Ans.  12  rd. 

3.  There  are  43560  sq.  ft.  in  1  A. :  what  is  each  side  of 
a  square,  containing  1  A,  ,',  A,   i  A  ? 

^ws.  208.71+ft.;  147.58-[-ft.;  and  104.35-f-ft. 

4.  A  man  has  2  fields;  10  A.  and  12^  A. :  find  the  side 
of  a  sq.  field  equal  in  area  to  both.  Ans.  60  rd. 

EXTRACTION    OF    THE    CUBE    ROOT. 

Art.  292.  To  extract  the  cube  root  of  a  number,  is 
to  resolve  it  into  three  equal  factors ;  or,  to  find  a 
number  which,  when  multiplied  by  itself  twice^  will  pro- 
duce the  given  number. 

Thus,  4  is  the  cube  root  of  64,  because  4X4X4  =  04. 
Koots.     1,      2,       3,      4,        5,        6,        7,        8,        9. 
Cubes.    1,       8,     27,     64,     125,     216,     343,     512,     729. 

Rkview. — 291.  ITow  find  tho  side  of  a  scjuaro  equal  in  area  to  any 
given  surface?     292.  What  is  it  to  extract  tho  cube  root  of  a  number  ? 


CUBE  ROOT.  287 

Art.  293.  From  Art.  278,  it  follows  that  the  cube  root 
of  a  number  expresses  the  side  of  a  cube  whose  solid 
contents  are  equal  to  the  given  number. 

Hence,  extracting  the  cube  root,  is  finding  the  side  of  a 
cube  when  its  solid  contents  are  known ;  or,  arranging  a  given 
number  of  cubes,  so  as  to  form  the  largest  cube  possible. 

Art.  294.  From  Art.  280,  it  follows  that, 

The  cube  root  of  1  is      1 ; 

The  cube  root  of  1000  is    10 ; 

The  cube  root  of  1000000     is  100;  and  so  on  :  hence, 
The  cube  root  of  a  number  between  1  and  1000,  consists  of 
one  figure  ;  between  1000  and  1000000,  of  two  figures;  between 
1000000  and  1000000000,  of  three,  &c. :  hence. 

Rule  for  Pointing. — If  a  dot  (.)  be  placed  over  every  3d 
figure  of  any  given  number,  beginning  with  units,  the  number 
of  dots  will  denote  the  number  of  figures  in  the  cube  root. 

Art.  295.  1.  Extract  the  cube  root  of  13824;  or, 
suppose  13824  cubic  blocks,  each  1  in.  long,  1  in.  wide, 
and  1  in.  thick,  are  to  be  arranged  in  the  form  of  a  cube. 

SoLU.— First   separate  operation.        13824(20+4 

the   given    number    into  8000      ='^4 


periods,  by  placing  dots  I Root 

over  the  4  and  3.  20X20X3  =  1200    5824 

The  root  will  consist  of  20 X    4X3=   240 
two  ^gnres.  4X    4        =      16 

We  next  find  that  the  ~1~t^   f^ft94. 

largest    cube    contained  = 

in  13  (thousand)  is  8  (thousand),  the  cube  root  of  which  is  2  (tens), 
which  place  on  the  right,  as  in  extracting  the  square  root. 

Subtract  the  cube  of  2  (tens),  which  is  8  (thousand),   from  the 
given  number,  and  5824  remain. 

While  solving  this  example  by  figures,  attend  to  arranging  the 
cubic  blocks.     After  finding  that  the  cube  root  of  the  given  number 


Eetiew. — 292.  Of  what  niimbens  are  the  nine  digits  the  cube  roots  ? 
293.  What  does  the  cube  root  of  a  number  express  ? 

294.  What  the  cube  root  of  a  number  between  1  and  1 000  ?  Why  ?  Of 
a  number  between  1000  and  1000000  ?  Why  ?  What  the  rule  for  pointing  ? 


288 


RAY'S  PRACTICAL   ARITHMETIC. 


will  contain  two  places  of  figures,  (tens  and  units,)  and  thai,  the 
figure  in  the  tens'  place  is  2,  form  a  cube,  A,  20  (2  tens)  inches 
long,  20  in.  wide,  and  20 in.  high;  this  cube  will  contain  (Art.  92; 
20X20X20  =  8000  cu.  in.;  take  this  sum  from  the  whole 
number  of  cubes,  and  5824  cu.  in.  are  left,  which  correspond  to 
the  number  5824  in  the  numerical  operation. 

It  is  obvious  that  to  increase  the  figure  Fig.  1. 

A,  and  at  the  same  time  preserve  it  a 
cube,  the  length,  breadth,  and  hight,  must 
each  receive  an  equal  addition. 

Then,  since  each  side  is  20  in.  long, 
square  20,  which  gives  20  X  20  =  400,  for 
the  number  of  sq.  in.  in  each  face  of  the 
cube ;  and  since  an  addition  is  to  be  made 
to  three  sides,  multiply  the  400  by  3,  which 
gives  1200  for  the  number  of  square  inches  in  the  3  sides. 

This  1200  is  called  the  trial  divisor;  because,  by  means  of  it, 
the  thickness  of  the  additions  may  be  determined. 

By  examining  Fig.  2  (or  the  blocks,  see  Note  6),  it  will  be  seen, 
that  after  increasing  each  of  the  three  sides  equally,  there  will  be 
required  3  oblong  solids,  C,  c,  c,  of  the  same  length  as  each  of 
the  sides,  and  whose  thickness  and  hight  are  each  the  same  as  the 
additional  thickness;  and  also  a  cube,  D,  whose  length,  breadth, 
and  hight,  are  each  the  same  as  the  additional  thickness. 

Hence,  the  solid  contents  of  the  first  three  rectangular  solids, 
the  three  oblong  solids,  and  the  small  cube,  must  together  be  equal 
to  the  remaining  cubes  (5824). 

Now  find  the  thickness  of  the  additions.  It  will  always  be 
something  less  than  the  number  of  times  the  trial  divisor  (1200)  is 
contained  in  the  dividend  (5824).  Fio.  2, 

By  trial  we  find  1200  is  contained  4 
times  in  5824.  Place  the  4  in  the  quo- 
tient, and  proceed  to  find  the  contents 
of  the  different  solids:  these  added 
together,  make  the  number  to  be  sub- 
tracted, called  the  subtrahend. 

The  solid  contents  of  the  first  three 
additions,  B,  B,  B,  are  found,  (Art.  93) 
by  multiplying  the  number  of  sq.  in.  in  the  face  by  the  thickness  j 


CUBE   ROOT.  289 

now  there  are  400  sq.  in.  in  the  face  of  each,  and  400  X  3  =  1200  sq. 
in.  in  one  face  of  (lie  three;  then  multiplying  by  4,  (the  thickness,) 
gives  4800  cu.  in.  for  their  contents. 

The  solid  contents  of  the  three  oblong  solids,  C,  c,  c,  are  found 
(Art.  93;  by  multiplying  the  number  of  sq.  in.  in  the  face  by  the 
thickness;  now  there  are  20  X  4  ^=  80  sq.  in.  in  one  face  of  each, 
and  80  X  3  =  240  sq.  in.  in  one  face  of  the  three  ;  then  multiplying 
by  4,  (the  thickness,)  gives  960  cu.  in.  for  their  contents. 

Lastly,  find  the  contents  of  the  small  cube,  D,  by  multiplying  its 
length  (4)  by  its  breadth  (4),  and  that  product  by  the  thickness  (4); 
this  gives  4  X  4  X  4  =  64  cu.  in.  additions. 

If  the  solid  contents  of  the  several     B  B    B  =4800cu.  in. 
additions  be  added  together,  their  sum,      q     q    (,  __    ogQ 
6824  cu.   in.,    will    be   the   number   of  D  =      64 

small   cubes   remaining   after  forming  

the  first  cube,  A.  ^^"^  5824 

Hence,  when  13824  cu.  in.  are  arranged  in  the  form  of  a  cube, 
each  side  is  24  in. ;  that  is,  the  cube  root  of  13824  is  24. 

In  finding  the  solid  contents  of  the  additions,  in  each  case  tha 
last  multiplier  is  the  thickness. 

To  produce  the  same  result  more  conveniently,  find  the  area 
of  one  face  of  each  of  the  additional  solids,  then  the  sum  of  the 
areas,  (as  in  the  numerical  operation,)  and  multiply  it  by  the 
common  thickness. 

The  sum  of  the  areas  of  one  face  of  each  of  the  additional 
solids,  is  termed  the  complete  divisor.     Thus, 

In   the  preceding  operation,  1456  is  the  complete  divisor. 

Note. — As  the  1st  figure  of  the  root  is  always  in  the  tens'  place 
with  regard  to  the  2d,  annex  to  it  a  cipher  before  it  is  squared;  or, 
omit  the  cipher,  and  multiply  the  square  by  300  instead  of  3. 

Eeview. — 295.  How  obtain  the  1st  figure  of  the  root  ?  "Why  square  it  ? 
Why  multiply  by  3  ?     What  is  the  product  called  ?     Why  ? 

295.  How  obtain  the  2d  figure  of  the  root  ?  Why  multiply  the  1st  figure 
by  the  2d  ?  Why  multiply  their  product  by  3  ?  Why  square  the  2d  figure 
of  the  root  ?     How  find  the  subtrahend  ?     What  is  the  complete  divisor  ? 

295,  Note.  Why  is  a  cipher  annexed  to  the  1st  figure  of  the  root  before 
squaring?  If  the  cipher  is  omitted,  what  must  be  done?  What  if  the 
cipher  is  omitted  in  multiplying  the  1st  figure  of  the  root  by  the  2d? 

3d  Bk.         19 


290 


RAY'S  PRACTICAL  ARITHMETIC. 


For  the  same  reason,  annex  a  cipher  to  the  figure  first  obtained, 
before  multiplying  it  by  the  2d  (the  thickness) : 

Or,  omit  the  cipher,  and  multiply  by  30  instead  of  3. 

Thus,  20  X  20  X  3  =  1200  2  X  2  X  300  =  1200 

20  X    4  X  3  =   240     Or,  2  X  4  X    30  ^   240 

4X    4        = 16  4X4  =16 

1456. 
^^2.  What  is  the  cube  root  of  1728? 
3.  Find  the  cube  root  of  413493625. 


Ans.  12. 


OPERATION.  413493625(745  Ans. 

343 


7X7X300  =  14700 
7X4X  30=  840 
4X4  =       16 


15556 


74X74X300=1642800 

74X    5X    30=     11100 

5X    5  =  25 


1653925 


70493  =  dividend. 


62224  =  subtrahend. 


8269625  =  dividend. 


8269625  =  subtrahend. 


Explanation. — By  the  rule  for  pointing  (Art,  294),  the  root 
will  contain  3  figures.  Find  the  1st  and  2d  figures  of  the  root  as 
in  the  preceding  examples.  Then  consider  74  as  so  many  tens, 
and  find  the  3d  figure  in  the  same  manner  as  the  2d  was  obtained. 

4.  Find  the  cube  root  of  515.849608 

OPERATION. 


515.849608(8.02  Ans. 
512 


8X8X300  =  19200 


80  X  80  X  300  - 

1920000 

80  X 

2X 

HO- 

4800 

2X 

2 

' — 

4 

1924804 


3849 


3849608 


3849608 


Exp. — After  obtaining  the  1st  figure,  and  bringing  down  the  2d 
period,  we  find  the  trial  divisor  is  not  contained  in  the  dividend; 


CUBE    ROOT.  291 

therefore,  place  a  0  in  the  root,  and  bring  down  another  period. 
Hence,  the  cube  root  of  a  decimal  is  found  in  the  same  manner 
as  that  of  a  whole  number,  the  periods  being  reckoned  both  ways 
from  the  decimal  point. 

Art.  296.  TO  EXTEACT  THE  CUBE  ROOT, 

Rule. — 1.  Separate  the  given  number  into  periods  of  Z places 
each,  by  placing  a  dot  over  the  units,  a  dot  over  the  thousands, 
and  so  on.  (The  left  period  often  has  only  one  or  two  figures.) 

2.  Find  the  greatest  cube  in  the  left  period,  and  place  its 
root  on  the  right,  as  in  division.  Subtract  the  cube  of  the  root 
from  the  left  period,  and  to  the  remainder  bring  down  the  next 
period  for  a  dividend. 

3.  Square  the  root  found,  and  multiply  it  by  300 /or  a  trial 
divisor.  Find  how  many  times  this  divisor  is  contained  in  the 
dividend,  and  write  the  result  in  the  root.  Multiply  the  last 
figure  of  the  root  by  the  rest,  and  by  30;  square  the  lastfgure 
of  the  root,  and  add  these  two  products  to  the  trial  divisor; 
the  sum  will  be  the  complete  divisor. 

4.  Multiply  the  complete  divisor  by  the  lastfgure  of  the  root, 
and  subtract  the  product  from  the  dividend;  to  the  remainder 
bring  down  the  next  period  for  a  new  dividend,  and  so  proceed 
until  all  the  periods  are  brought  doivn. 

Notes. — 1.  When  the  product  of  the  complete  divisor  by  the 
last  figure  of  the  root  is  larger  than  the  dividend,  the  figure  of  the 
root  must  be  diminished. 

2.  After  bringing  down  all  the  periods,  if  there  be  a  remainder, 
the  operation  may  be  continued  by  annexing  periods  of  ciphers. 

3.  If  the  divisor  is  not  contained  in  the  dividend,  write  a  cipher 
in  the  root,  and  bring  down  another  period  for  a  new  dividend. 

4.  When  there  are  decimals  in  the  given  number,  separate  them 
into  periods  by  placing  dots  over  the  tenths,  ten-thousandths,  and 
80  on.    The  reasons  for  this  are  similar  to  those  in  Art.  286,  Ex.  4. 

5.  To  extract  the  cube  root  of  a  common  fraction,  reduce  it  to 
its  lowest  terms;  then,  if  both  terms  are  perfect  cubes,  extract 
the  cube  root  of  each;  but,  if  either  term  is  an  imperfect  cube, 
reduce  the  fraction  to  a  decimal,  and  then  extract  the  root. 

Sl^^  Every  Teacher  should  have  a  set  of  cubical  blocks. 


292 


RAY'S    PRACTICAL    ARlTHiMETIC. 


To  Teachers. — Instead  of  finding  the  subtrahend  by  the  rule,  it  may 
be  obtained  by  finding  separately  the  contents  of  each  solid,  then  adding 
the  whole  together.  This  method,  in  connection  with  the  blocks,  is  best 
adapted  to  give  a  clear  idea  of  the  nature  of  the  operation. 

What  is  the  Cube  Root  of, 


ANSWERS. 

ANSWERS 

5. 

91125? 

45. 

15. 

53.157376?   3.76 

6. 

195112? 

58. 

16. 

.199176704?  .584 

7. 

912G73? 

97. 

17. 

216?              6, 
343  '               7* 

8. 

1225043? 

107. 

18. 

2  7  4  4?            14 

Eb 5  9  '          J  y • 

9. 

13312053? 

237. 

19. 

48778  ?         29 
118638*          "3'5* 

10. 

102503232? 

468. 

20. 

5iM?      If- 

11. 

529475129? 

809. 

21. 

2?     1.25992-h 

12. 

958585256  ? 

986. 

22. 

9?     2.08008-h 

13. 

14760213677? 

2453. 

23. 

200?    5.84803-f 

U. 

128100283921 ? 

5041. 

24. 

91?      2.092-1- 

25.  The  contents  of  a  cubical  cellar  are  1953.125  cu. 
ft.  :  find  the  length  of  one  side.  Ans.  12.5ft. 

26.  In  1  cu.  ft.,  how  many  3  in  cubes  ?  Ans.  64. 

27.  How  many  cubical  blocks,  each  side  of  which  is 
one-quarter  of  an  inch,  will  fill  a  cubical  box,  each  side 
of  which  is  2  inches  ?  Ans.  512. 

28.  Find  the  difference  between  half  a  solid  foot,  and 
a  solid  half  foot.  Ans.  648cu.  in. 


29.  Find  the  side  of  a  cubical  mound  equal  to  one 
288  ft.  long,  216  ft.  broad,   48  ft.  high.  Ans.  144  ft. 

30.  The  side  of  a  cubical  vessel  is  1  foot :  find  the 
side  of  another  cubical  vessel  that  shall  contain  3  times 
as  much.  Ans.  17.3064-in. 


Review. — 296.  What  the  rule  for  extracting  the  cube  root?  Notes. 
When  the  subtrahend  is  larger  than  the  dividend,  what  is  required  ? 

296.  W^hen  there  is  a  remainder,  how  continue  the  operation  ?  TIow 
proceed  when  there  are  decimals  in  the  given  number  ?  How  extract 
the  cube  root  of  a  couimnn  friictiun  ? 


ARITHMETICAL    PROGRESSION.  293 

Art.  297.  It  is  a  known  principle,  that  sphere.^  are  to 
each  other  as  the  cubes  of  their  diameters  ;  and  that 

All  similar  solids  are  to  each  other  as  the  cubes  of  their 
corresponding  sides. 

Hence,  the  solid  contents,  or  weight  of  two  similar  solids,  have 
to  each  other  the  same  ratio  as  the  cubes  of  their  like  paints. 

31.  A  metal  ball  Gin.  in  diameter  weighs  321b. :  what 
is  the  weight  of  one  of  the  same  metal,  whose  diameter 
is  3  in.  ?  Ans.  4  lb. 

32.  If  the  diameter  of  Jupiter  is  11  times  that  of  the 
earth,  how  many  times  larger  is  it?  Ans.  1331. 

Art.  298.  THE    CUBE    BOOT    BY    FACTORING. 

The  cube  root  of  any  perfect  cube  may  be  extracted 
bi/  resolving  the  given  number  into  its  prime  factors,  and 
multiplying  together  one  of  each  three  equal  factors. 

1.  Find  the  cube  root  of  216.      .     .     Arm.  3X2=6. 

2.  Find  the  cube  root  of  27X64.     .     .     .     Ans.  12. 

3.  Find  the  cube  root  of  125X343.      .     .     ^ns.  35. 


XXV.  ARITHMETICAL  PKOGRESSION. 

Art.  299.  An  Arithmetical  Progression,  or  Series,  is  a 
series  of  numbers  which  increase  or  decrease,  by  a  common 
difference.  If  the  series  increase,  it  is  called  an  increasing 
series  ;  if  it  decrease,  a  decreasing  series. 

Thus,     1,     3,     5,     7,     9,  11,  &c.,  is  an  increasing  series. 
20,  17,  14,  11,     8,     5,  &c.,  is  a    decreasing  series. 

The  numbers  forming  the  series  are  called  terms;  the  first 
and  last  terms  are  the  extremes  ;  the  other  terms,  the  means. 

Review. — 297.  "What  ratio  have  the  solid  contents  of  two  similar 
bodies  ?    298.  How  extract  the  cube  root  of  a  perfect  cube  by  factoring  ? 

299.  "What  is  an  arithmetical  progression  ?  "When  is  the  series  increas- 
ing ?  Decreasing  ?   Give  examples.   What  are  the  extremes  ?  The  means  ? 


294  RAY'S  PRACTICAL    ARITHMETIC. 

Art.  300.  In  every  arithmetical  series,  five  things  are 
considered: 

1st,  the  first  term  ;  2d,  the  last  term ;  3d,  the  common 
difference;  4th,  the  number  of  terms;  5th,  the  sum  of 
all  the  terms. 

CASE    I. 

Art.  301.  To  find  the  last  term,  when  the  first  term, 
the  common  difference,  and  the  number  of  terms  are  given. 

1.  I  bought  10  yd.  of  muslin,  at  3cts.  for  the  1st  yd., 
7cts.  for  the  2d,  llcts.  for  the  3d,  and  so  on,  with  a 
com.  difference  of  4cts. :  what  did  the  last  yd.  cost? 

Solution. — To  find  the  cost  of  the  second  yard,  add  4  cts.  once 
to  the  cost  of  the  first;  to  find  the  cost  of  the  third,  add  4  cts.  twice 
to  the  cost  of  the  first;  to  find  the  cost  of  the  fourth,  add  4  cts.  three 
times  to  the  cost  of  the  first,  and  so  on.     Hence, 

To  find  the  cost  of  the  tenth  yard,  add  4 cts.  nine  times  to  the  cost 
of  the  first;  but  9  times  4  cts.  are  36  cts.,  and  3cts.-)-3()  cts.:^39cts., 
the  cost  of  the  last  yard,  or  last  term  of  the  progression. 

2.  The  first  term  of  a  decreasing  series  is  39 ;  the  com. 
diff.  4;  the  number  of  terms  10 :  find  the  last  term. 

Solution. — In  this  case,  4  must  be  subtracted  9  times  from  39, 
which  will  give  three  for  the  last  term.     Hence,  the 

Kule  for  Case  I. — Multiply  the  common  difference  hy  the 
number  of  terms  less  one ;  if  an  increasing  series,  add  the 
product  to  the  \st  term:  if  a  decreasing  series,  subtract  the 
product  from  the  \st  term  :  the  result  will  be  the  required  term. 

3.  Find  the  last  term  of  an  increasing  progression :  the 
first  term  2;  the  common  difference  3;  and  the  number 
of  terms  50.  Ans.  149. 

4.  I  bought  100  yd.  muslin,  at  9  cts.  for  the  1st  yard., 
14  cts.  for  the  2d,  and  so  on,  increasing  by  the  com.  dif- 
ference 5  cts. :  find  the  cost  of  the  last  yd.       Ans.  $5.04 

5.  What  is  the  54th  term  of  a  decreasing  series,  the 
1st  term  140,  and  com.  diff.  2?  Ans.  34. 

6.  A  lends  $200  at  simple  interest,  at  8  ^  per  annum : 

Revikw. — :<00.  What  five  things  uro  considered  in  every  series  ? 


ARITHMETICAL  PROGRESSION.  295 

at  the  end  of  the  1st  year  $216  will  be  due;  at  the  end 
of  the  2d  year,  $232,  and  so  on :  what  sum  will  be  due 
at  the  end  of  20  years?  Ans.  $520. 

7.  What  is  the  99th  term  of  a  decreasing  series,  the 
1st  term  329,  and  com  diff.  |?  Ans.  2431. 

CASE    II. 

Art.  302.  To  find  the  common  difference,  when  the 
extremes  and  the  number  of  terms  are  given. 

1.  The  first  term  of  a  series  is  2,  the  last  20,  and  the 
number  of  terms  7:  what  the  com.  difF.? 

Solution. — The  difference  of  the  first  and  last  terms  is  always 
equal  to  the  com.  diff.  multiplied  by  the  number  of  terras  less  one 
(Art.  301);  therefore, 

If  the  difference  of  the  extremes  be  divided  by  the  number  of 
terms  less  one,  the  quotient  will  be  the  com.  diff.     Hence,  the 

Bule  for  Case  II. — Divide  the  difference  of  the  extremes  hy 
the  number  of  terms  less  one;  the  quotient  will  be  the  com.  diff. 

2.  The  extremes  are  3  and  300;  the  number  of  terms 
10:  find  the  com.  difi".  Ans. '6^. 

3.  A  travels  from  Boston  to  Bangor  in  10  da.;  he  goes 
5  mi.  the  first  day,  and  increases  the  distance  traveled  each 
day  by  the  same  number  of  miles ;  and  on  the  last  day  he 
goes  50 mi.:  find  the  daily  increase.  Ans.  5 mi. 

Art.  303  It  is  obvious  that  if  the  difference  of  the 
extremes  be  divided  by  the  com.  diff.,  the  quotient, 
increased  by  unity  (1),  will  be  the  number  of  terms. 

1.  The  extremes  are  5  and  49;  the  com.  diff.  4:  find 
number  of  terms.  Ans.  12. 

CASE    III. 

Art.  304.  To  find,  the  SUM  of  all  the  terms  of  the  series^ 
when  the  extremes  and  number  of  terms  are  given. 

'Revikw.— 301.  What  is  Case  1?  What  is  the  Rule  for  Case  1? 
802.  What  is  Case  2  ?    What  is  the  Rule  for  Case  2  ? 


1, 

3, 

5, 

7, 

9, 

11, 

11, 

9, 

7, 

5, 

3, 

1, 

296  RAY'S   PRACTICAL   ARITHMETIC. 

1.  Find  the  sum  of  6  terms  of  the  series  whose  first  term 
is  1..  and  last  term  11. 

Solution. — Tlie  series  is  . 
The  order  inverted  is  .  . 
The  sum  is 12,     12,     12,     12,     12,     12. 

Since  the  two  series  are  the  same,  their  sum  is  twice  the  first 
geries.  But  their  sum  is  obviously  as  many  times  12,  (the  sum  of 
the  extremes),  as  there  are  terms.     Hence,  the 

Rule  for  Case  III. — Multiply  the  sum  of  the  extremes  hy  the 
number  of  terms ;  Jialf  the  product  will  he  the  sum  of  the  series. 

2.  The  extremes  are  2  and  50 ;  the  number  of  terms,  24 : 
find  the  sum  of  the  series.  Ans.  624. 

3.  How  many  strokes  does  the  hammer  of  a  clock  strike 
in  12  hours?  Ans.  78. 

4.  Find  the  sum  of  the  first  ten  thousand  numbers  in  the 
series,    1,   2,   3,   4,    5,  ka.  Ans.  50005000. 

5.  Place  100  apples  in  a  right  line,  3  yd.  from  each  other, 
the  first,  3  yd.  from  a  basket :  Avhat  distance  will  a  boy  travel 
who  gathers  them  singly  and  places  them  in  the  basket  ? 

Ans.  17  mi.  380  yd. 

C.  A  traveled  one  day  30 mi.,  and  each  succeeding  day  a 
quarter  of  a  mile  less  than  on  the  preceding  day :  how  far  did 
be  travel  in  30  days  ?  Ans.  791 J  mi. 

7.  A  body  falling  by  its  own  weight,  if  not  resisted  by  the 
air,  would  descend  in  the  1st  second  a  space  of  16  ft.  1  in.  ; 
the  next  second,  3  times  that  space ;  the  3d,  5  times  that 
space ;  the  4th,  7  times,  <fcc.  ;  at  that  rate,  through  what  space 
would  it  fall  in  1  min.  ?  Ans.  57900  ft. 


XXVI.  GEOMETRICAL  PROGRESSION. 

Art.  305.  A  Geometrical  Progression^  or  Series^  is  a  series 
of  numbers  increasing  by  a  common  multiplier^  or  decreasing 
by  a  common  divisor.     Thus, 

1,       3,       9,     27,     81,  is  an  increasing  geometric  series. 
48,     24,     12,       0,       3,  is  a    decreasing  geometric  series. 


Kkview.— G03.  How  find  tho  number  of  terms,  when  tho  extremes  and 
common  difference  are  given  ? 


GEOMETRICAL   PROGRESSION.  297 

The  common  multiplier  or  com.  divisor,  is  called  the  ratio. 
In  the  1st  of  the  above  series,  the  ratio  is  3 ;    in  the  2d,  2. 

The  numbers  forming  the  series  are  the  terms ;  the  first  and 
last  terms  are  extremes;  the  others,  means. 

Art.  306 .  In  every  geometric  series,  5  things  are  considered : 
1st,  the^rs^  term;    2d,   the  last  term;    3d,   the  number  of 
terms  ;  4th,  the  ratio ;  5th,  the  sum  of  all  the  terms. 

CASE    I. 

Art.  307.  To  find  the  last  term,  when  the  first  term^  the 
ratio.,  and  the  number  of  terms  are  given. 

1.  The  first  term  of  an  increasing  geometric  series  is  2  ;  the 
ratio  3 ;  what  is  the  5th  term  ? 

Solution. — The  first  term  is  2;  the  second,  2X3;  the  third, 
2X3X3;  the  fourth,  2X3X3X3;  and 

The  fifth,  2X3X3X3X3=2X34  =2X81  =162.  Ans. 

Observe  that  each  term  after  the  first,  consists  of  the  first  term 
multiplied  by  the  ratio  taken  as  a  factor  as  many  times  less  one,  as 
is  denoted  by  the  number  of  the  term.  Thus,  ih^  fifth  term  consists 
of  2  multiplied  by  3  taken /owr  times  as  a  factor.  But  3,  taken  4 
times  as  a  factor,  is  (Art.  277)  the  4th  power  of  3.     Hence, 

The  fifth  term  is  equal  to  2,  multiplied  by  the  4th  power  of  3. 

2.  The  first  term  of  a  decreasing  geometric  series  is  192  ;  the 
ratio  2 ;  what  is  the  fourth  term  ? 

Solution.— The  2d  term  is  192 -f- 2;  the  3d  is  192  divided 
by  2X2;  the  4th  is  192  divided  by  2X2X2;  that  is, 
192 -r- 2"^=  192-^-8  =  24,  Ans.  The  required  term  is  found  by 
dividing  the  first  term  by  the  ratio  raised  to  a  power  whose  ex- 
ponent is  1  less  than  the  number  of  the  term.     Hence,  the 

Rule  for  Case  I. — Raise  the  ratio  to  a  power  (Art.  279) 
whose  exponent  is  one  less  than  the  number  of  terms. 

If  the  series  be  inci^easing, multiply  the  1st  term  by  this  power, 
and  the  product  will  be  the  last  term ;  if  decreasing,  divide  the 
1st  term  by  the  poioer.  and  the  quotient  will  be  the  last  term. 

Review.— 304.  What  is  Case  3?  What  is  the  Rule  for  Case  3? 
805.  What  is  a  geometrical  series  ?  Give  examples.  What  the  ratio  ? 
What  the  extremes  ?   The  means  ?  306.  What  five  things  are  considered  ? 


298  RAYS   TRACTICAL  ARITHMETIC. 

Note. — In  finding  high  powers  of  the  ratio,  the  operation  may 
often  be  shortened  by  observing  that  the  product  of  any  two  powers 
of  a  number,  will  give  that  power  of  the  number  which  is  denoted 
by  the  sum  of  their  exponents.     Thus, 

The  third  power  multiplied  by  the  fourth  power,  will  produce  the 
seventh  power. 

23x24  =8X10=3  128  =  2'^. 

3.  The  first  term  of  an  increasing  series  is  2;  the  ratio,  2; 
the  number  of  terms,  13  :  find  the  last  term.      Ans.  8192. 

4.  The  first  term  of  a  decreasing  series  is  262144;  the 
ratio,  4 ;  number  of  terms,  9 :  find  the  last  term.      Ans.  4. 

5.  The  first  term  of  an  increasing  series  is  10;  the  ratio,  3: 
what  the  tenth  term?  Ans.  196830. 

6.  What  the  3r)th  term  of  an  increasing  series,  whose  first 
term  is  1,  and  ratio,  2?  Ans.  17179869184. 

7.  Find  the  35ih  term  of  an  increasing  series,  the  1st  term, 
1;  ratio,  3.  Ans.  10677181699660569. 

CASE   II. 

Art.  308.  To  find  the  sum  of  all  the  terms  of  a  geometric 
series. 

1.  To  obtain  a  General  Rule,  let  us  find  the  sum  of  5  terms 
of  the  geometric  series,  whose  1st  term  is  4,  and  ratio  3. 

Solution. — Write  the  terms  of  the  series  as  below ;  then  multi- 
ply each  terra  by  the  ratio,  and  remove  the  product  one  term 
toward  the  right:   thus, 

4+12  +  36  +  108  +  324  =  sum  of  the  series. 

12  +  36  +  108  +  324  +  972  =  sum  X  3. 

Since  the  upper  line  is  once  the  sum  of  the  series,  and  the  lower 
three  times  the  sum,  their  difference  is  tioice  the  sum;  hence. 

If  the  upper  line  be  subtracted  from  the  lower,  and  the  remainder 
divided  by  2,  the  quotient  will  be  the  sum  of  the  series. 

Performing  this  operation,  we  have  972 — 4:=:9G8;  which,  di- 
vided by  2,  the  quotient  is  484,  the  required  sum. 

In  this  process,  972  is  tlie  product  of  the  greatest  term  of  the 
given  series  by  the  ratio;  4  is  tlie  least  term,  and  the  divisor  2,  is 
equal  to  the  ratio  less  one. 

Review. — S07.  "What  is  Case  1  ?  What  the  Rule  ?  Note.  How  aro 
high  powora  of  tbo  ratio  most  easily  found  ? 


PERMUTATION.  299 

Rule  for  Case  II. — Multiply  the  greatest  term  hy  the  ratio ; 
from  the  product  subtract  the  least  term^  and  divide  the  remainder 
hy  the  ratio  less  1 ;   the  quotient  will  he  the  sum  of  the  series. 

Note. — When  a  series  is  decreasing,  and  the  number  of  terms 
infinite,  the  last  term  is  naught.  In  finding  the  sum  by  the  rule, 
observe  that  the  ratio  is  greater  than  1. 

2.  The  first  term  is  10;  the  ratio,  3 ;  the  number  of  terms, 
7:  what  is  the  sum  of  the  series?  Ans.  10930. 

3.  A  gave  to  his  daughter  on  New  Year's  day  $1  ;  he 
doubled  it  the  first  day  of  every  month  for  a  year :  what  sum 
did  she  receive  ?  Ans.  $4095. 

4.  I  sold  1  lb.  of  gold  at  1  ct.  for  the  1st  oz.,  4  for  the  2d, 
16  for  the  3d,  &c.:  what  the  sum?  Ans.  $55924.05 

5.  A  sold  a  house  having  40  doors,  at  10  cts.  for  the  1st  door, 
20  for  the  2d,  40  for  the  3d,  and  so  on:  how  much  did  he 
receive?  Ans.  $109951162777.50 

6.  B  bought  a  horse  at  1  ct.  for  the  1st  nail  in  his  shoes, 
3  for  the  2d,  9  for  the  3d,  ka. :  what  was  the  price,  there  being 
32  nails  ?  Ans.  $9265100944259.20 

7  Find  the  sum  of  an  infinite  series,  the  greatest  term  .3; 
the  ratio,  10;  that  is,  of  y^{j+yoo~l~To%0' *^*^-  ^'^^-  \- 

8.  Find  the  sum  of  an  infinite  series,  greatest  term  100; 
ratio  1.04  Ans.  2600. 

9.  The  sum  of  the  infinite  series  i,  ^,  ^V)  ^^-  Ans.  ^. 
10.  The  sum  of  the  infinite  series  g,  5,  g,    ^^c.           Ans.  1. 


XXYII.    PERMUTATION. 

Art.  309.  Permutation  teaches  the  method  of  finding  in 
how  many  different  positions  any  given  number  of  things  may 
be  placed. 

Thus,  the  two  letters  a  and  h  can  be  placed  in  ttco  positions, 
ah  and  ba;  but  if  we  add  a  third  letter  c,  three  positions  can  be 
made  with  each  of  the  two  preceding :  thus. 

Cab,  acb,  abc,  and  cba,  bca,  bac,  making  2X3=6  positions. 

By  taking  a.  fourth  letter  d,  four  positions  can  be  made  out 
of  each  of  the  six  positions,  making  6X4=24  in  all. 


300  RAY'S  PRACTICAL  ARITHMETIC. 

Rule  for  Permutation.  —  Multiply  together  the  num- 
bers^ 1,  2,  3,  &c.^from  1  to  the  given  number ;  the  last  product 
will  be  the  required  result. 

1.  In  how  many  different  ways  may  the  digits  1,2,  3,  4, 
and  5  be  placed  ?  Ans.  1 20. 

2.  What  number  of  changes  may  be  rung  on  12  bells  ? 

Ans.  479001600. 

3.  What  time  will  8  persons  require  to  seat  themselves  dif- 
ferently every  day  at  dinner,  allowing  365  days  to  the  year? 

Ans.  llOyr.   170 da. 

4.  Of  how  many  variations  do  the  26  letters  of  the  alpha- 
bet admit  ?  Ans.  403291461 126605635584000000. 


XXVIII.   MENSURATION. 

To  Teachers. — As  this  short  article  on  Mensuration  is  intended  for 
pupils  who  may  not  have  an  opportunity  of  studying  a  more  extensive 
course,  only  the  more  useful  parts  are  presented. 

The  definitions  and  illustrations  are  given  in  plain  and  familiar  terms, 
not  with  a  view  to  mathematical  precision. 

Art.  310.   DEFINITIONS. 

1.  An  Angle  is  the  inclination  of  two  straight  lines  meetinj; 
in  a  point,  which  is  called  the  Veutex.  It  is  the  degree  of  the 
opening  of  the  lines. 

2.  When  one  straight  line  stands  on  another  so 
that  it  makes  with  it  two  equal  angles,  each  of 


Right 
angle, 


Right        these  angles  is  a  Right  Angle;  and  the  straight 
au-ie.       line  which  stands  on  the  other  is  said  to  be  Peu- 


PENDicuLAE  to  it,  or  at  KiGUT  Angles  to  it. 

Obtuso     /Acute  3.  An  Odtuse  Angle  is   greater  than  a 

angle.     /  angle.       right  angle:  and  an  Acute  Angle  is  less  than 
g jj- jy        a  right  angle. 

Note. — An  angle  is  named  by  3  letters,  the  middle  one  being 
placed  at  the  vertex,  and  the  otlier  two  on  the  lines  which  form 


Review.— 308.  What  is  Case  2  ?  What  the  Rule  ?  Note.  What  is 
the  last  term  of  a  decreasing  series  of  which  the  number  of  terms  is 
infinite  ?     809.  What  is  Permutation  ?     What  the  Rule  ? 


MENSURATION. 


301 


the  angle.     In  the  diagram  the  obtuse  angle  is  called  the  angle 
A  C  B,  and  the  acute  angle,  the  angle  A  C  D. 

4.  Parallel    straight    lines  are  everywhere 

equally  distant  from  each  other. 


5.  A  Surface  has  length  and  breadth,  without  thickness. 

6.  A  Plane  is  a  surface,  in  which,  if  any  two  points  be  taken, 
the  straight  line  joining  them  wnll  be  wholly  in  the  plane. 

7.  A  Figure  is  a  portion  of  surface  inclosed  by  one  or  more 
boundaries. 

8.  If  a  ficrure  has  equal  sides,  it  is  E-qui-lat^-er-al  ;   if  it  has 
e(iual  angles,  E-qui-an^-gu-lar. 

9.  A  Triangle  is  a  figure  bounded  by  3 
straififht  lines.  The  side  on  which  the  triano-le 
stands,  is  the  Base.  The  perpendicular  hight 
is  the  shortest  distance  from  the  base  to  the  op- 
posite angle.  Thus,  A  B  C  is  a  triangle  ;  A  C 
is  the  base,  and  B  D  the  perpendicular  hight.  " 

10.  A  Quad-ri-lat'-er-al  is  bounded  by  4  straight  lines. 

11.  A  Polygon  is  bounded  by  more  than  4  straight  lines, 

12.  A  Par-al-lel^-o-gram  is  a  quadrilateral 
whose  opposite  sides  are  parallel. 

Thus,  M  N  O  P  is  a  parallelogram. 

13.  A  Rect^-an-gle  is  a  quadrilateral  whose 
opposite  sides  are  parallel;  its  angles  right 
angles.     Thus,  11 E  C  T  is  a  rectangle. 

14.  A  Square  is  a  quadrilateral  whose  sides 
are  equal  to  each  other;  its  angles  right  angles. 
Thus,  S  Q  U  A  is  a  square. 

15.  A  Rhom^-bus  is  a  quadrilateral  whose  sides  are  equal  to 
each  other ;  its  angles,  not  right  angles.     See  Fig.  Def.  12. 

16.  A  Trap^-e-zoid  is  a  quadrilateral  having 
only  two  sides  parallel.  Thus,  Z  O I  D  is  a 
trapezoid ;  the  sides  Z  D  and  0 1  being  parallel. 

17.  A  Tra-pe^-zium  is  a  quadrilateral  having 
no  two  sides  parallel  tn  each  other. 

Thus,  T  R  A  P  is  a  trapezium. 


302  RAY'S   PRACTICAL    ARITHMETIC. 

18.  A  Di-AG''-o-NAL  is  a  line  joining  two  angles  of  a  figure 
not  next  to  each  other.  Thus,  6  U  (Fig.  Dof.  14)  is  a  diagonal 
of  the  square. 

19.  A  Circle  is  a  figure  bounded  by  a  curve 
line,  called  the  circumference^  every  part  of 
which  is  equally  distant  from  a  point  within, 
called  the  center.  A  Diameter  is  a  straight 
line  passing  through  the  center  and  terminated 
both  ways  by  the  circumference.  A  Radius  is 
a  straight  line  drawn  from  the  center  to  the 
circumference;   it  is  half  the  diameter. 

Thus,  A  D  B  E  is  the  circumference ;  A  B  the  diameter ; 
A  F  or  B  F  the  radius. 

20.  A  Tangent  is  a  straight  line  which  touches  the  circum- 
ference only  in  one  point,  called  the  point  of  contact 

Thus,  T  1*  is  a  tangent. 

21.  An  Arc  of  a  circle  is  any  part  of  the  circumference,  as 
A  F.  A  Chord  is  a  straight  line  joining  the  extremities  of 
an  arc. 

22.  Men.suration  is  the  art  of  finding  the  surface,  and  also 
the  solid  contents  of  bodies. 

23.  The  Area  of  a  figure  is  the  surface  which  it  contains. 
The  quantity  of  this  surface  is  denoted  by  the  number  of  times 
it  contains  a  given  surface  called  the  measuring  unit. 

The  MEASURING  UNIT  for  surfiices  is  a  square  surface,  whose 
side  is  some  one  of  the  common  measures  of  length,  such  as  a 
square  inch,  a  square  foot,  <fcc.     iSee  Arts.  87,  88,  and  89. 

MEASUREMENT    OF    SURFACES. 
Art.  311.  To  find  the  superCcial  contents  or  Area  of 
a  Parallelogram,  Rectangle,  Square,  or  Rhombus, 

Rule. — Multiply  the  length  hy  the  perpendicular  breadth, 
the  product  will  be  the  area. 

Note. — The  learner  nrmst  recollect  (Art.  27G)  thnt  feet  in  length 
multiplied  by  feet  in  breadth,  produce  square  feet;  and  the  same 
of  the  other  denominations  of  lineal  measure. 

1.  ITow  many  square  feet  in  a  floor  17  feet  long  and  15  feet 
wide?  Ans.  !;255sq.  ft 


MENSURATION.  303 

2.  Find  the  sq.  ft  in  a  board  2  ft.  3  in.  wide  and  12  ft 
6  in.  long.  Ans.  28.125  =  28^  sq.  ft 

3.  The  sq.  ft  in  a  board  15  in.  wide  and  16  ft  long. 

Ans.  20  sq.  ft. 

4.  How  many  sq.  ft  in  a  board  1  ft  2  in.  in  mean  breadth, 
12  ft  6  in.  long?       Ans.  14 sq.  ft  84  sq.  in.,  or  H/^^q-  ft- 

5.  At  $1.50  per  sq.  ft,  what  cost  a  marble  slab;  the 
length  5ft  Tin.;  breadth  1  ft  lOin.  ?         Ans.  $15,354+ 

6.  How  many  acres  of  land  in  a  parallelogram;  the  lenjith, 
120  rd. ;  the  breadth,  84  rd.  ?  Ans.  63  A. 

7.  How  many  acres  in  a  square  field,  each  side  of  which  is 
65  rd.  ?  Ans.  26  A.  1  R.  25  P. 

8.  How  many  acres  in  n  field  in  the  form  of  a  rhombus  ; 
each  side  measures  35  rd  ,  the  perpendicular  distance  across 
it,  16  rd.?  Ans.  3  A.  2  R. 

9.  Each  side  of  the  base  of  a  pyramid  is  693  ft.  long :  how 
many  acres  does  it  cover?  Ans.  11  A.  4  P. 

10.  Find  the  difference  between  a  floor  30  ft  sq.,  and  two 
others  each  15  ft  sq.  Ans.  450  sq.  ft 

n.  If  a  room  is  10  ft  long,  how  wide  must  it  be  to  contain 
80  sq.  ft  ?     (See  Art.  90.)  Ans.  8  ft 

12.  A  board  is  10 inches  wide:  what  must  be  its  lenscth  to 
contain  lOsq.  ft?  A7is.  l2ft 

13.  How  many  yd.  of  carpet  l^yd.  wide,  will  cover  a  floor 
6yd.  long,  5yd.  wide?  Ans.  20yd. 

14.  How  many  yd.  of  carpet  1  ^  yd.  wide,  will  cover  a  floor 
21ft.  3  in.  long,  13  ft  6 in.  wide?  Ans.  25^ yd. 

15.  How  many  yd.  of  flannel  |  yd.  wide,  will  line  3  yd.  of 
cloth,  l^yd.  wide?  Ans.  6yd. 

PIasterers\  Pavers\  Painters^  and   Carpenters     Work. 

Art.  312.  Several  kinds  of  artificers'  work  are  measured 
by  the  preceding  rule. 

Plasterers',  Pavers',  and  Painters'  work,  is  computed  in  sq. 
yards:  Glaziers'  work,  by  the  sq.  ft,  or  by  the  pane: 

Carpenters'  and  Joiners'  work,  some  parts  by  the  sq.  yard ; 
other  parts  by  the  Square,  which  contains  100  sq.  ft 

1.  How  many  square  yards  in  a  ceiling  25  ft  9  in  long, 
and  21  ft  3  in.  wide?  Ans.  60 sq.  yd.  7  sq.  ft-f 


I 

k 


304  RAYS    PRACTICAL    ARITHMETIC. 

2.  At  20ct8.  a  sq.  yd.,  what  will  it  cost  to  plaster  a  ceil- 
ing 22  ft.  7  in    long,  13  ft.  11  in.  wide  ?  Ans.  $6,984+ 

3.  A  room  is  20  ft.  6  in.  long,  10  ft.  3  in.  broad,  10  ft.  4  in. 
high:  how  many  yd.  of  plastering  in  it,  deducting  a  fireplace 
Oft.  Sin.  by  4ft  2  in. ;  a  door  7  ft.  by  4  ft.  2  m.,  and  two 
windows,  each  6  ft.  by  3  ft.  3  in.  ?    Ans.  110  sq  yd.  Sj^  sq.  ft 

4.  A  room  is  20ft.  long,  14  ft.  Gin.  broad,  and  10ft  4  in. 
high  :  what  will  the  coloring  of  the  walls  cost,  at  27  cts.  per 
sq.  yd.,  deducting  a  fireplace  4ft  by  4  ft.  4  in.,  and  two  win- 
dows, each  0  ft  by  3  ft  2  in.  ?  Ans.  $19.73 

5.  At   18  cts.    per  sq.  yd.,   find   the  cost  of  paving  a  walk 

35  ft  4  in.  long,  8ft  3  in.  broad.  ^«6'.  $5.83 

6.  What  will  it  cost  to  pave  a  rectangular  yard,  21  yd.  long, 
and  15  yd.  broad,  in  which  a  footpath,  5ft  3  in.  wide,  runs 
the  whole  length  of  the  yard ;  the  path  paved  with   flags,  at 

36  cts.  per  sq.  yd.,  and  the  rest  with   bricks,  at  24  cts.  per 
sq.  yd.?  Ans.  $80.01 

7.  At  10  cts.  a  sq.  yd.,  what  the  cost  to  paint  the  walls  of  a 
room  75  ft.  6  in.  in  compass,  12  ft  6  in.  high  ?  Ans.  $10,486+ 

8.  A  house  has  3  tiers  of  windows,  7  in  a  tier :  the  hight 
of  the  first  tier  is  6  ft  11  in.  ;  of  the  2d,  5  ft  4  in.  ;  the  3d, 
4  ft  3  in.  ;  each  window  is  3  ft.  G  in.  wide :  what  cost  the 
glazing,  at  16  cts.  per  sq.  ft  ?  Ans.  $64.68 

9.  A  floor  is  .36  ft  3  in.  long,  16  ft  6  in.  wide ;  what  will  it 
cost  to  lay  it,  at  $3  a  square  ?  Ans.  $17,943  + 

10.  A  room  is  35ft  long,  and  30ft  wide:  what  will  the 
flooring  cost,  at  $5  per  square,  deducting  a  fireplace  Oft  by  4 
ft  G  in.,  and  a  stairway,  8  ft  by  10  ft  6  in.  ?     Ans.  $46.95 

11.  At  $3.50  per  square,  what  cost  a  roof  40  ft  long,  the 
rafters  on  each  side  18ft  6  in.  long?  Ans.  $51.80 

Art.  313.  To  find  the  Area  of  a  Triangle. 

Kule. — MuUijjly  the  base  by  the  perpendicular  hiyht,  and 
take  half  the  product  for  the  area. 

Or,  when  the  sides  are  given,  the  following  Rule  : 
Ist   Add  the  three  sides  together.,  and  take  half  the  sum. 
2d.   From  the  half  sum  take  the  3  sides  severally. 
3d.  Multiply  the  half  sum  and  the  3  remainders  together,  and 
extract  the  square  root  of  the  product,  which  gioes  the.  area. 


MENSURATION.  805 

1.  Find  the  area  of  the  triangle,  E  F  G  H, 

the  base,  F  II,  is  15  feet;  the  perpendicuhir  E 

hight,  G  E,  12  feet.  Ajis.  UOsq.  ft 

2.  The  contents  of  a  triangular  space,  the 
base  44  rd.,  perpendicular  hight  18rd, 

Ans.  2  A.  1  R.  36P. 

3.  How  many  acres  in  a  triangular  field ;  the  base  80  rd. ; 
perpendicular  hight,  67  rd.  ?  Ans.  16  A.  3  R. 

Note. — The  area  of  any  field  or  piece  of  land  may  be  found  by 
dividing  it  into  triangles,  and  measuring  the  base  and  perpendic- 
ular hight  of  each  triangle  thus  formed. 

4.  What  cost  the  glazing  of  a  triangular  skylight,  at  12  cts. 
per  sq.  ft.,  the  base,  12  ft.  6  in.,  the  perpendicular  hi^ht,  16  ft 
9in.  ?  Ans.  $12,561 

5.  Find  the  area  of  a  triangle,  the  sides  being  13,  14,  and 
15  ft  Ans.  848q.  ft 

6.  The  area  of  a  triangle,  the  sides  2,  3,  and  4  feet  respect- 
ively. Ans.  2.9047375+sq.  ft 

Art.  314.  To  find  the  Area  of  a  Trapezoid. 

Hule. — Multiply  the  sum  of  the  parallel  sides  hj  the  perpen- 
dicular breadth ;  take  half  the  product. 

1.  The  parallel  sides  of  a  trapezoid,  F  C  G  D, 
are  35  and  26  inches;  its  breadth  llin. ; 
required  the  area.  Ans.  335^  sq.  in.         c 

2.  A  field  is  the  form  of  a  trapezoid ;  one  of  the  parallel 
sides  is  25  rd.,  the  other  19  rd. ;  the  width  32  rd. :  how  many 
acres  in  it?  Ans.  4 A.  1  R.  24 P. 

Art.  315.  To  find  the  Circumference  of  a  Circle. 

WHEN    THE    DIA:METER   IS   GIVEN. 

Rule. — Multiply  the  diameter  6?/  3.1416,  the  product  will  he 
the  circumference. 

1.  The  diameter  A  B  of  the  circle  A  D  B  E  ^ 
is  48  feet :  what  is  the  circumference  ? 

Ans.  150.7968  ft         a£^^^B 

2.  The  diameter  of  a  wheel  is  4  feet :  find 
the  circumference.      Ans.  12  ft  6.7968  in. 

3d  Bk.  20 


/ 


306  RAYS   PRACTICAL   ARlTHxMETIC. 

3.  What  is  the  circumference  of  the  earth,  the  mean  di- 
ameter being  7912.4  mi.  ?  Ajis.  24857.59584  mi. 

Art.  316.  To  find  the  Diameter   of  a  Circle,   when 

THE   circumference    IS   GIVEN. 

Rule. — Divide  the  circumference  by  3.1416,  the  quotient 
will  he  the  diameter. 

1,  The  circumference  of  a  circle  is  15 feet:  what  is  the 
diameter  ?  Ans.  4  ft.  9.2i)54-in. 

2.  If  the  girt  of  a  tree  is  12  feet  5  inches,  what  its  thick- 
ness or  diameter  ?  ^7is.  3  ft  11.428+in. 

Art.  317.  To  find  the  Area  of  a  Circle. 

Kule. — Multiply  the  diameter  by  the  circumference,  and  take 
one-fourth  of  the  product.  Or,  Multiply  the  square  of  the 
diameter  by  .7854;  or,  for  greater  accuracy^  by  .785398 
Or,  Multiply  the  square  of  the  radius  by  3.1416 

1.  Find  the  area  of  a  circle,  the  diameter  being  42  feet. 

Alls.  1385.4456  sq.ft. 

2.  Find  the  area  of  a  space  on  which  a  horse  may  graze, 
when  confined  by  a  cord  7^  rods  long,  one  of  its  ends  being 
fixed  at  a  certain  point  Ans.  1 A   16.715P. 

Art.  318.  To  find  the  Diameter  of  a  Circle,  when 

THE  Area  is  given. 

Unle. — Divide  the  area  by  .7854 ;  the  square  root  of  the 
quotient  will  be  the  diameter. 

1.  The  area  of  a  circle  is  962.115:  what  its  diameter  and 
circumference?  ^7is.  diam.  35 :  circum.   109.950 

2.  What  length  of  halter  will  fasten  a  horse  to  a  post  in  the 
center  of  an  aero  of  grass,  so  that  he  can  graze  upon  the  1  A. 
and  no  more  ?  Ans.  7.13G4+rd.,''or  117  ft  9-i-in. 

Art.  319.  MEASTTREMENT    OF    BODIES    OR    SOLIDS. 

Definitions. — 1.  A  Body  or  Solid,  has  length,  breadth,  and 
thickness  or  depth.  p 

2.  A  Prism  is  a  solid  whoso  ends,  or 
biiHos,  are  parallel ;  its  sides,  parallelo- 
grams.    Such  a  body  is  termed  a  Right 


MENSURATION. 


307 


Prism  when  each  of  its  bases  is  perpendicular  to  its  other  sides ; 
and  it  is  Triangular,  Quadrangular,  &c.,  according  as  its  base 
is  a  triangle,  quadrangle,  &c.     Thus,  P  is  a  triangular  prism. 

3.  A  Par-al-lel-o-pi'-ped  is  a  prism  whoso 
bases  and  also  its  other  sides  are  parallelo- 
grams.    Thus,  B  is  a  parallelopiped. 

4.  A  parallelopiped  is  right  when  each  of 
its  faces  is  a  rectangle.  A  common  chest,  a 
bar  of  iron,  brick,  &c.,  are  instances  of  right 
parallelopipeds.  When  each  face  of  a  right 
parallelopiped,  as  A,  is  a  square,  it  is  termed 
a  cube.    A  cube  has  6  equal  square  faces. 


5.  A  Cylinder  is  a  round  prism,  having 
circles  for  its  ends.  Thus,  C  is  a  cylinder, 
of  which  the  line  E  F  passing  through  the 
centers  of  both  ends,  is  called  the  axis. 


6  A  Pyramid  is  a  solid  having  any  plane 
figure  for  a  base,  and  its  sides  triangles, 
whose  vertices  meet  in  a  point  at  the  top, 
called  the  vertex  of  the  pyramid.  A  pyra- 
mid is  Triangular,  Quadrangular,  &c., 
according  as  its  base  is  a  triangle,  quad- 
rangle, &c.  Thus,  A  is  a  triangular  pyramid. 


7.  A  body  which  has  a  circular  base,  and 
tapers  uniformly  to  a  point  named  the  ver- 
tex, is  called  a  Cone.  The  axis  of  a  cone  is 
a  line  passing  through  the  vertex  and  the 
center  of  the  base.  Thus,  C  is  a  cone  of 
which  B  V  is  the  axis. 


8.  A  Frustum  of  any  body,  as  a  pyramid  or  cone,  is  what 
remains  when  the  top  is  cut  off  by  a  plane  parallel  to  the  baset 


808 


RAY'S   PRACTICAL   ARITHMETIC. 


9.  A  Globe,  or  Sphere,  is  a  body  of  such 
a  figure,  that  all  points  of  the  surface  are 
equally  distant  from  a  point  within,  called 
the  center. 

The  diameter  of  a  sphere,  is  a  line  pass- 
ing through  the  center,  and  terminated  both 
ways  by  the  surface.  The  radius  of  a  sphere 
is  a  line  drawn  from  the  center  to  the  surface. 

Thus,  A  B  is  a  diameter:  CA  a  radius; 
C  being  the  center  of  the  sphere. 

10.  The  HiGHT  or  altitude  of  a  solid,  is  a  line  drawn  from 
its  vertex  or  top,  perpendicular  to  its  base. 

11.  The  Contents  or  Solidity  of  a  body,  is  the  space  within 
it  The  magnitude  of  this  space  is  expressed  by  the  number  of 
times  it  contains  a  given  space  called  the  measuring  unit. 

12.  The  Measuring  Unit  for  solids,  is  a  cube  whose  base  is 
the  measuring  unit  for  surfaces;  as  a  cu.  in.,  cu.  ft.,  &c. 


Art.  320.  To  find  the  Solid  Contents  of  a  Parallelopiped. 

Rule. — Multiply  the  lengthy  breadth,  and  depth  together: 
the  product  will  he  the  solid  contents. 

1.  Find  the  solid  contents  of  a  parallelopiped :  the  length, 
12  ft. ;  breadth,  3  ft  3  in. ;  depth,  4  ft  4  in.     Ans.  169  cu.  ft 

2.  The  solid  contents  of  a  rectangular  stone :  the  length,  6  ft. ; 
breadth,  2  ft  6  in. ;  depth,  1  ft  9  in.  Ans.  26^  cu.  ft 

3.  A  block  of  marble,  in  the  form  of  a  parallelopiped,  is  in 
length,  3  ft  2  in.,  breadth,  2ft  8  in, ;  depth,  2ft  6in. :  what 
its  cost,  at81cts.  per  cu.  ft?  Ans.  $17.10 

4.  How  many  solid  feet  in  a  box  4  ft  10  in.  long,  2  ft  1 1  in. 
broad,  and  2  ft  2  in.  deep  ?  Ans.  30  cu.  ft^O^  G^'  4'^'. 

Art.  321.  The  principles  of  the  preceding  rule  are  applied 
to  the  measurement  of 

MASONS'   AND    BRICSLAYEES'    WORK. 

Masons'  work  is  measured  by  the  solid  foot,  or  by  the  perch, 
which  is  IG^ft  long,  18  in.  broad,  1  ft  deep; 

And  multiplying  these  numbers  together,  shows  that  a  perch 
contains  24  jj,  or  !24.75  cu.  ft 


MENSURATION.  309 

To  find  the  number  of  Perches  in  any  Wall,  or  solid  body. 

Hule. — Find  the  contents  in  cubic  feet^  by  multiplying 
together  the  lengthy  breadth^  and  depth ;  then  divide  by  24.75 
to  obtain  the  contents  in  perches. 

1.  How  many  perches  in  a  wall  97  ft.  5  in.  long,  18  ft.  Sin. 
high,  2  ft  3  in.  thick?  Ans.  161.6+  P. 

2.  In  a  wall  53  ft  6  in.  long,  12  ft  3  in.  high,  2ft  thick, 
how  many  perches  ?  Ans.  52.95+  P. 

3.  What  cost  a  wall  53  ft  6  in.  long,  12  ft  6  in.  high,  2ft 
thick,  at  $2.25  a  perch?  "       Ans.  $121.59+ 

4.  How  many  bricks  in  a  wall  48  ft  4  in.  long,  16  ft  6  in. 
high,  1  ft.  6  in.  thick,  20  bricks  to  the  cu.  ft  ?     Ans.  23925. 

5.  At  $5,875  per  thousand  bricks,  allowing  20  bricks  to 
wall  the  solid  foot,  what  will  it  cost  to  build  a  wall  320  ft  long, 
6ft  high,  15  in.  thick?  Ans.  $282. 

6.  How  many  bricks  each  Sin.  long,  4  in.  wide,  2.25  in. 
thick,  will  be  required  for  a  wall  120  ft  long,  8ft  high,  1  ft 
6  in.  thick  ?  Ajis.  34560. 

7.  What  the  cost  of  building  a  wall  240  ft.  Ions;,  6  ft  hiorh, 

3  ft.  thick,  at  $3.25  per  1000  bricks,  each  brick  9  in.  long, 

4  in.  wide,  2  in.  thick  ?  Ans.  $336.96 

Art.  322.    To    find  the  Solid   Contents    of  a    Prism, 

OR  OF  A  Cylinder. 

Rule. — Find  the  area  of  the  base,  and  multiply  it  by  the 
pei'pendicular  hight,  the  product  will  be  the  solid  contents. 

1.  Each  side  of  the  base  of  a  triangular  prism  is  2  in. ;  its 
length  14 in.:  find  the  contents.   A.  i/588=24.2487+cu.  in. 

2.  Find  the  contents  of  a  cylinder  12  ft  long,  the  diameter 
of  each  end,  4  ft  Ans.  150.7968  cu.  ft 

3.  The  cu.  in.  in  a  bu.,  each  end  184  in.  in  diameter,  depth  8  in. 

^ns.  2150.4252  cu.  in. 

4.  If  the  bu.  contain  2150.4  cu.  in.,  what  are  the  solid 
contents  of  a  cylindrical  tub  6  ft  in  diameter  and  8  ft.  deep  ? 

Ans.  181.764  bu. 

5.  How  many  bu.  in  a  box  15  ft  long,  5  ft  wide,  4  ft  deep? 

Ans.  241+  bu. 

6.  How  many  bu.  in  a  box  12  ft.  long,  3  ft  wide,  5ft  deep? 

Ans.  144.6+  bu. 


310  RAY'S   PRACTICAL  ARITHMETIC. 

Art.  323.  To  find  the  Solid  Contents  of  a 
Pyra^iid  or  of  a  Cone. 

Kule. — Multiply  the  area  of  the  base  by  the  perpendicular 
hight,  and  take  one-third  of  the  product. 

1.  Find  the  solid  contents  of  a  square  pyramid,  the  base, 
5ft  each  side;  perpendicular  hight,  21ft      Ans.  175cu.ft 

2.  The  solid  contents  of  a  cone,  base  10ft.  in  diameter; 
perpendicular  hight,  15  ft  ^?is.  392.7  cu.  ft 

3.  The  diameter  of  the  base  of  a  conical  glass-house  is  37  ft. 
8  in. ;  the  altitude,  79  ft  9  in. :  what  space  is  inclosed  ? 

Ans.  29622.0227+cu.  ft 

4.  A  sq.  pyramid  is  477ft  high;  each  side  of  its  base  720 
ft :  find  the  contents  in  cu.  yd.  Ans.  3052800  cu.  yd. 

5.  How  often  can  a  conical  cup  9  in.  deep,  I^in.  diameter, 
be  filled  from  a  gal.  or  231  cu.  in.  ?  Ans.  43*57 -ftimes. 

Art.  324.  To  find  the  Solid  Contents  of  the  FrvUSTUM 
OF  A  Pyramid,  or  of  the  Frustum  of  a  Cone. 

Rule. — 1st  Find  the  area  or  surface  of  each  end  by  the 
j)receding  rules.  2d.  Find  the  area  of  the  mean  base  by  multi- 
plying the  areas  of  the  upper  and  lower  bases  together.,  extracting 
the  square  root  of  the  product.  3d.  Add  together  the  areas  of 
the  tipper,  the  lower.,  and  the  mean  base;  multiply  their  sum  by 
one-third  of  the  altitude^  the  product  will  be  the  solid  contents, 

1.  Find  the  solid  contents  of  a  block  with  square  ends,  each 
side  of  the  lower  base  3  ft ;  and  of  the  upper  base  2  ft :  the 
altitude,  12  ft  Ans.  7Gcu.  ft 

2.  The  length  of  the  frustum  of  a  sq.  pyramid  is  18  ft.  8  in. ; 
the  side  of  its  greater  base  27in. ;  that  of  its  lesser  base,  10 
in.:  what  the  contents?  Ans.  Gl. 2283950 -f-cu.  ft. 

3.  Find  the  contents  of  a  glass  in  the  form  of  the  frustum 
of  a  cone;  the  diameter  at  the  mouth  2Jiin.;  at  the  bottom, 
1  in.;  the  depth  5  in.  Ans.  12.70275  cu.  in. 

Art.  325.  To  find  the  Solid  Contents  of  a  Globe. 

Rule. — Multiply  the  cube  of  the  diameter  by  .5230 

1.  What  are  the  solid  contents  of  3  globes,  their  diameters 
being  13,  15,  nnd  30  inches,  respectively? 

Ans.  1150.3492;   1707.15;  and  14137.2  cu.  in. 


MENSURATION.  311 

Art.  326.  To  find  the  Area  of  the  Surface  of  a  body 

BOUNDED  BY  PLANE  SURFACES. 

Hule. — Find  the  area  of  the  surfaces  separately;  then  add. 

To  FIND  THE  Area  of  the  Curved  Surface  of  a  Right  Cone. 

Kule. — Multiply  the  circumference  of  the  hose  by  the  slant 
hight,  and  take  half  the  prodvxit 

To  find  the  Surface  of  a  Globe. 
Bnle. — Multiply  the  square  of  the  diameter  by  3.1416 

1.  Each  side  of  the  base  of  a  triangular  pyramid  is  5  ft. 
4  in. ;  its  slant  hight,  from  the  vertex  to  center  of  each  side  of 
the  base  7  ft.  6  in. :  find  the  area  of  its  surface.    Ans.  60  sq.ft. 

2.  What  is  the  convex  surface  of  a  cone,  whose  side  is 
25  ft.,  and  diameter  of  the  base  8^  ft.  ?     Ans.  333.795  sq.  ft 

3.  Find  the  area  of  the  curved  surface  and  base  of  a  right 
cone,  the  slant  hight  4  ft  7  in. ;  the  diameter  of  its  base  2  ft 
1 1  in.  Ans.  27.679895+sq.  ft 

4.  If  the  earth  be  a  perfect  sphere  7912  mi.  in  diameter, 
what  its  superficial  contents  ?     Ans.  196663355.7504  sq.mi 

Art.  327.  GAUGING 

Is  tbe  method  of  finding  the  contents  of  any  regular  vessel, 
in  gallons,  bushels,  barrels,  &c. 

When  the  vessel  is  in  the  form  of  a  cube  or  parallelepiped, 
apply  this 

Rule. —  Take  the  dimensions  in  inches  and  multiply  the 
length.,  breadth^  and  depth  together; 

This  product  divided  &?/  23 1 ,  will  give  the  contents  in  wine 
gal. ;  or,  divided  by  2150.4,  will  give  the  contents  in  bu. 

1.  How  many  wine  gallons  in  a  trough,  10  ft  long,  5  ft  wide, 
and  4ft  deep?  Ajis.  1496+gaL 

2.  How  many  bushels  in  a  box,  12  ft  long,  6  ft.  wide,  and 
10  ft  deep  ?  Ans.  578.57+bu. 

Art.  328.  To  find  the  Contents  of  a  Cistern,  both  ends 
Circular,  the  Upper  and  Lower  Diaiieters  equal. 

Rule. — Take  all  the  dimensions  in  inches;  then  square  the 
diameter,  multiply  this  by  the  hight,  and  this p)roduct  by  .7854; 
this  will  give  the  contents  in  cubic  inches,  and   this  divided 


312  RAY'S  PRACTICAL  ARITHMETIC. 

by   231    will  give   the  contents  in  wine  gallons,   which  may 
he  reduced  to  barrels  by  dividing  by  31.5. 

Note. — Since  .7854-^-231=.0034,  therefore,  when  required  to 
multiply  by  .7854  and  divide  the  product  by  231,  shorten  the 
operation  (Art.  61),  by  multiplying  at  once  by  .0034. 

1.  How  many  barrels  in  a  cistern,  the  diameter  beinci;  4  ft, 
the  depth  6  ft?  Ans.  17.9+bl. 

2.  How  many  barrels  in  a  cistern,  the  diameter  6  ft.,  and 
depth  9  ft.  ?  Ans.  00.43+  bL 

Art.  329.    To  find  the  Contents  of  a  Cistern,  both  ends 
Circular,  and  Diameters  Unequal. 

Rule. — Having  the  dimensions  in  inches,  multiply  iogeihe? 
the  diameters  of  the  two  ends ;  to  the  product  add  one-third  of 
the  square  of  the  difference  between  these  diameters : 

Then  multij^ly  this  sum  by  the  hight,  and  the  product  by 
.7854 :  the  result  will  be  cu.  in.,  which  reduce  as  in  Art  328. 

1.  What  the  contents,  in  wine  gallons,  of  a  cistern,  the  upper 
diameter  40  in. ;  lower  diameter  30  in. ;  depth  50  in.  ? 

Ans.  209.G6+  gal. 

2.  The  contents  in  bl.,  of  a  cistern,  the  upper  diameter  7  ft 
G  in. ;  lower  diameter  10  ft ;  depth,  12  ft  6  in.  ?   Ans.  179^  bl. 

Art.  330   .   To  find  the  Contents  of  a  C.vsk  or  Barrel. 

When  the  staves  are  straight  from  the  bung  to  each  end, 
consider  the  cask  as  the  frustums  of  two  equal  cones  and 
find  the  contents  by  the  rule.  Art  324. 

When  the  staves  are  curved,  find  the  mean  diameter  of  the 
cask,  by  adding  to  the  head  diameter,  two-thirds  of  the  diScr- 
cnce  between  the  bung  and  head  diameters ;  or, 

If  the  staves  are  but  little  curved,  add  six-tenths  of  the 
difference;  and, 

Having  the  mean  diamet<!r,  find  the  solidity  in  the  samo 
manner  as  that  of  a  cylinder.  Art  322. 

Since  multiplying  the  square  of  half  the  mean  diameter 
by  3.1410  is  the  same  as  to  multiply  its  smuirc  by  .7854;  and 
multiplying  by  .7854  and  dividin<^  by  231,  (Art.  328,  Note,) 
is  the  same  as  to  multiply  by  .0034 ; 

Therefore,  to  find  the  contents  of  a  cask,  when  its  dimcnsionB 
are  in  inches,  apply  the  following: 


MENSURATION.  313 

J{,^e. Multiply  the  square   of  the  mean  diameter  hy  the 

lengthy  and  that  product  by  .0034;   the  result  will  be  wine  gal. 

1.  How  many  gallons  in  a  cask,  the  staves  much  curved,  the 
buno-  diameter  40  in.,  head  diameter  31  in.,  length  50  in.  ? 

Ans.  232.73  gal. 

2.  Find  the  contents  of  a  cask,  the  staves  nearly  straight, 
bung  diameter  32  in.,  head  diameter  30  in.,  length,  40  in. 

Ans.  132.38+  gaL 

Art.  330   .    MISCELLANEOUS    EXAMPLES. 

1.  A  rectangular  field  is  15 rods  long:  what  must  be  its 
breadth  to  contain  one  acre?  Ans.  10|rd. 

2.  How  many  cubic  feet  in  a  room  24ft.  lonir,  18ft.  Gin. 
wide,  10ft  Tin.  high?  Ans.  4699 cu.  ft 

3.  The  area  of  a  circle  is  1  sq.  ft :  what  its  diameter  ? 

Ajis.  13.5405-f  in. 

4.  The  solid  contents  of  a  globe  are  I  cu.  ft. :  what  tho 
diameter?  ^7ts.  14.8884+ in. 

5.  The  sides  of  a  triangle  arc  30,  40,  and  50  feet :  required 
the  area.  Ans.  66|  sq.  yd. 

6.  How  many  sq.  ft.  in  a  plank  12  ft  6  in.  lon^;  one  end 
15  in.  broad,  the  other  11  in.  ?  Ans.  13J|sq.  ft 

7.  Two  circles,  10  and  10  ft.  in  diameter,  have  the  same 
center:  what  their  difference  of  area?  Ans.  122.5224 sq.  ft. 

8.  "What  will  it  cost  to  line  a  rectangular  cistern,  6  ft  long, 
2  ft  broad,  2  ft.  6  in.  deep,  with  sheet  lead,  at  4  cts.  a  lb. ;  allow- 
ing 81b.  of  lead  to  each  sq.  ft  of  surface?  Ans.  $16.64 

9.  At  25  cts.  a  bushel,  what  will  oats  cost  to  fill  a  bin  5  ft 
long,  4  ft.  wide,  4  ft  deep  ?  Ans.  %  1 6.07+ 

10.  What  is  the  area  of  a  circle,  of  which  the  circumfer- 
ence is  1448  ft  ?     Ans.  3  A.  3  R.  12  P.  25  sq.  yd.  8+  sq.  ft 

11,  Required  the  surface  of  a  cube,  each  side  being  37  in. 

Ans.  8214  sq.  in. 

Art.  331.  MECHANICAL  POWERS.— This  subject  pro- 
perly belongs  to  a  more  advanced  work,  and  will  bo  found 
appropriately  treated  in  "  Ray's  Higher  Arithmetic^ 


314  RAY'S    PRACTICAL    ARITHMETIC. 


Art.  332.— 100  Peomiscxjous  Questions. 

1.  The  sum  of  three  equal  numbers  ia  1236:  what  is  one 
of  the  numbers  ?  A7is.  412. 

2.  The  sum  of  two  equal  numbers,  less  225,  is  675 :  find 
jne  of  the  numbers.  Ans.  450. 

3.  There  are  four  equal  numbers,  whose  sum  divided  by  3, 
is  292:  find  one  of  them.  Ans.  219. 

4  What  cost  5  lb.  15  oz.  of  tea,  at  $1.20  per  lb.  ? 

Ans.  $7. 12 J 

5.  What  cost  13  bu.  3  pk  potatoes,  at  $1.45  per  bu.  ? 

Ans.  $19.93| 

6.  Two  men,  A  and  B,  purchased  a  farm  of  320  acres ;  A 
paid  $1000,  and  B  paid  $600 :  how  many  acres  should  each 
receive?  Ans.  A,  200;  B,  120  acres. 

7.  In  what  time  will  a  man,  walking  at  the  rate  of  3|  miles 
an  hour,  travel  42^  miles  ?  Ans.  1 1  hr.  20  min. 

8.  What  number  multiplied  by  1 1  will  =  14|  ?     A.  lOy^. 

9.  I  have  a  number  in  my  mind,  which  X  3,  =  81  less  than 
when  X  6  :  what  is  the  number  ?  Ans.  27. 

10.  A  man  bought  4yd.  of  cloth  at  $|  per  yd.,  and  10yd. 
at  $1  per  yd. :  he  paid  with  muslin  at  $^per  yd. :  how  many 
yards  were  required?  A7is.  111^ yd. 

11.  After  spending  |  of  my  money  and  1  of  what  was  left,  I 
had  $125  remaining  :  what  sum  had  1  at  first  ?  Ans.  $500. 

12.  Multiply  the  sum  of  2-^^  and  Ig  by  their  difforcnce,  ex- 
pressing the  product  decimally,  Ans.  3.30078125 

13.  I  was  married  at  the  age  of  21 :  if  I  live  19  yr.  longer,  I 
will  have  been  married  60 yr. :  what  is  my  age  ?     A7is.  62  yr. 

14.  Find  the  least  Com.  Mult  of  8,  12,  21,  36,  and  48,  and 
divide  it  by  the  greatest  Com.  Div.  of  65  and  143.  Ans.  77j^3. 

15.  How  many  French  meters,  each  39.371  Enixlish  inches, 
are  there  in  3  mi.  5  fur.   110yd.?  Ans.  5934.317+ 

16.  In  what  time  can  you  count  800000000,  at  the  rate  of 
250  a  min.,  counting  10  hr.  a  da.,  365  da.  to  the  yr.  ? 

Ans.  14  yr.  223  do.  3hr.  20  min. 

17.  Divide  12.625  by  KJf.  Ans.  .7575 


PaOMISCUOUS    QUESTIQNa  815 

18.  What  does  the  rent  of  a  house  amount  to  from  May  20, 
1854,  to  May  10,  1855,  at  $:;i50  per  year?     Aiis.  535243 Jg. 

19.  I  bought  an  equal  quantity  of  flour,  butter,  and  sugar, 
for  $47;  the  sugar  was  12  cts.,  the  butter  30  cts.,  and  the  flour 
5  cts.,  a  pound:  how  much  of  each  did  1  buy  ?    Aris.  1001b. 

20.  A  cistern  is  |  full  of  water ;  after  35  gal.  are  taken  out^ 
it  is  I  full:  how  many  gal.  will  it  contain?         Ans.  120 gal. 

21.  I  bought  60  barrels  of  flour  at  $5  a  barrel;  sold  23 
barrels  at  $4  a  bl. :  at  how  much  per  bL  must  1  sell  the  rest, 
to  gain  $51  on  the  whole?  Ans.  $7. 

22.  How  many  boxes  of  3qr.  131b.  each,  can  be  filled  from 
a  hhd.  of  sugar  containing  12  cwt.  1  qr.  7  lb.  ?  Ans.  14. 

23.  What  will  it  cost  to  gild  a  globe  10  inches  in  diameter, 
at  5  cents  per  square  inch?  A7is.  $15,708 

24.  If  1  ox  is  worth  8  sheep,  and  3  oxen  are  worth  2  horses, 
what  is  the  value  of  each  horse,  the  sheep  being  valued  at 
$2.50  each  ?  Ans.  $30. 

25.  If  I  of  $1  buy  ^  of  a  sheep,  and  |  of  a  sheep  be  worth 
-J^  of  an  ox,  what  will  10  oxen  cost?  Ans.  $200. 

26.  What  number  has  to  54  the  same  ratio  that  19  has 
to  9?  Ans.  114. 

27.  Two-thirds  of  the  ratio  of  |  to  |,  is  three  times  the  ratio 
of  3  to  what?  Ans.  1. 

28.  By  working  13  hr.  a  day,  a  man  can  perform  a  piece  of 
work  in  5^  days :  in  what  time  can  he  perform  it  by  working 
9  hr.  a  day  ?  Ans.  7^-^  da. 

29.  I  bought  501b.  of  tea  for  $40,  and  sold  it  so  as  to  clear 
$15:  had  1  purchased  $100  worth  of  tea,  and  sold  it  at  the 
same  rate,  what  sum  would  1  have  made  ?         Ans.  $37.50 

30.  A  clock  gains  7-i  min.  in  24  hr.  It  is  set  right  at  noon 
on  Monday:  what  will  be  the  time  by  it  at  6  o'clock  on  the 
following  Thursday  evening?  Ans.  6hr.  24|min. 

31.  If  7  men  can  mow  35  acres  of  grain  in  4  days,  how 
many  acres  will  10  men  mow  in  3i  days?        Ans.  43|A. 

32.  Bought  35|yd.  linen  at  $|  per  yd.,  and  sold  161  yd. 
at  $11  per  yd.,  and  the  rest  at  $|peryd. :  what  the  gain  by 
the  transaction  ?  Jns.  $5.67-|- 

33.  If  a  man  can  build  lOcu.  ft.  of  wall  in  an  hr.,  what 
length  of  wall,  5  ft.  high,  and  2  ft.  thick,  can  he  build  in  6  da., 
working  1 1  hr.  a  da.  ?  Ans.  06  ft. 


316  RAY'S   PRACTICAL  ARITHMETIC. 

34.  If  4  men  or  6  women  do  a  piece  of  work  in  20  da.,  how 
soon  can  3  men  and  5  women  do  H?  Ans.  12||da. 

35.  I  bought  40  yd.  of  cloth  at  the  rate  of  5  yd.  for  $6,  and 
60  yd.  more  at  the  rate  of  6yd.  for  $9:  I  sold  the  whole  at 
the  rate. of  5yd.  for  $7:  what  did  1  gain?  Ans.  $2. 

36.  From  a  vessel  containing  50  gal.  wine,  lOgaL  were 
drawn  off  and  the  vessel  filled  with  water:  if  10  gal.  more  be 
drawn,  how  many  gal.  pure  wine  will  remain  ?     Ans.  32  gal. 

37.  If  27  men  do  a  piece  of  work  in  14  days,  working  lOhr. 
a  day,  how  many  hours  a  day  must  24  boys  work,  to  perform 
it  in  45  da.,  2  boys  being  equal  to  I  man  ?  Ans.  7  hr. 

38.  A  ship  starts  at  noon  and  sails  west,  9  hours,  going 
16^  40^^  an  hour:  what  time  will  it  be  at  the  place  reached? 

Ans.  lOmin.  before  9. 

39.  If  10  men  in  10.2  days  of  9  hours  each,  dig  a  trench 
20.4  ft.  long,  3.3  ft.  wide,  1.5  ft.  deep,  in  how  many  days  of 
10  hours  each,  can  20  men  dig  one  40  ft  long,  4.5  ft.  wide, 
1.1  ft.  deep  ?  Ans.  9  da. 

40.  What  was  the  cost  of  an  article,  which,  when  sold  for 
$14,  paid  a  profit  of  20  fo  ?  Ans.  $1  If. 

41.  Two  men  hired  a  pasture  for  $45 ;  A  put  in  4  horses, 
and  B,  9  cows.  If  3  cows  equal  2  horses,  what  sum  must 
each  pay?  Ans.  A,  $18;  B,  $27- 

42.  Two  men.  having  started  at  the  same  time  to  travel 
toward  each  other  met  in  2^-  hr. ,  one  traveled  5  mi.  an  hr. 
faster  than  the  other,  and  both  together  traveled  35  mi. :  at 
what  rate  per  hr.  did  each  travel?         Ans.  4^-;  and  9^ mi. 

43.  I  sold  corn  for  $14.85,  and  lost  17^  fc  :  for  what  should 
I  have  sold  it  to  have  gained  12^  ^  ?  Ans.  $20.25 

44.  IIow  much  grain  must  I  take  to  mill,  so  that  I  shall  have 
2bu.  left  for  grinding,  after  paying  toll  at  the  rate  of  4  qt.  to 
the  bushel?  Ans.  2  bu.   1  pk.  IJ^qt 

45.  If  32  men  have  food  for  5  mon.,  how  many  must  leave, 
for  the  food  to  last  the  rest  8  mon.  ?  Ans.  12. 

46.  Received  $1009.29  for  a  note  having  60  da.  to  run,  dis- 
counted in  bank  at  0  %  :  how  much  should  I  have  received  for 
it,  discounted  by  true  discount,  at  12  ^  ?  Ans.  $1000. 

47.  I  bought  20  yd.  cloth  at  5  %  less  than  first  cost,  and 
sold  it  at  10  %  more  than  first  cost;  I  gained  $12:  what  was 
the  first  cost  per  yard  ?  Ans.  $4. 


PROMISCUOUS    QUESTIONa  317 

48.  Bought  a  metallic  plate  9  in.  square,  and  h  in.  thick, 
for  $3. '24;  what  should  1  pay  for  a  plate  of  the  same  metal 
1  ft.  2 in.  long,  11  in.  wide,  and  |in.  thick?         Ans.  $7.70 

49.  A  vessel  at  sea  has  120  persons  on  board,  and  provisions 
sufficient  to  last  3  mon. ;  they  take  from  a  wreck  60  persons 
more,  how  long  will  their  provisions  last  ?  Ans.  2  mon 

50.  A  fox  is  47 rd.  before  a  dog;  the  fox  runs  20 rd.  and 
the  dog  25  rd.  in  a  minute :  how  many  rods  must  the  dog  run 
to  catch  the  fox  ?  Ans.    235  rd. 

51.  For  what  sum  must  I  give  my  note  at  the  Bank  of  Boston, 
payable  in  4  months,  at  6  ^  discount,  to  obtain  $300  ? 

Ans.  $306.27+ 

52.  I  have  40  gallons  of  wine  worth  $1.50  per  gal,  which 
I  wish  to  reduce  to  $1.20  per  gal. :  how  much  water  must  I 
add?  Ans.  10  gal. 

53.  A  bank  charged  $1.26  for  discounting  my  note  at  60 
days :  what  was  the  amount  of  the  note,  the  rate  per  cent, 
being  6?  Ans.  $120. 

54.  A  hare  having  45rd.  the  start  of  a  dog,  can  run  25  rd. 
while  the  dog  runs  28  rd.  :  how  many  rods  must  the  dog  run 
to  catch  the  hare?  Ans.  420 rd. 

55.  Wheat  sold  for  $1.50  per  bu.,  pays  a  profit  of  -|  of  the 
cost :  if  sold  for  $2  a  bu.  what  fo  would  it  pay  ?     Ans.  60  fo. 

56.  I  spent  $260  for  apples  at  $1.30  per  bu. ;  after  retain- 
ing a  part  for  my  own  use,  I  sold  the  rest  at  40  fo  profit,  and 
cleared  $13  on  the  cost  of  the  whole:  how  many  bushels  did 
1  retain?  Ans.  50 bu. 

57.  Two  men  start  from  the  same  place,  and  travel  the  same 
road :  A  at  the  rate  of  8  miles  an  hour,  and  B  at  the  rate  of  10 
miles  an  hour;  A  starts  5  hours  before  B:  in  what  time  will 
B  overtake  A?  Ans.  20 hr. 

58.  A  can  mow  2  A.  in  3  da. ;  B,  5  A.  in  6  da. :  in  how  many 
da.  can  they  both  together  mow  9  A.  ?  Ans.  6  da. 

59.  A  and  B  together  do  a  piece  of  work  in  20  days:  in  what 
time  can  each  do  it  separately,  if  A  does  three  times  as  much 
as  B  ?  Ans.  A,  26§  da. ;  B,  80  da. 

60.  I  bought  two  equal  lots  of  oranges :  for  the  first,  1  gave 
5  cts.  for  2  oranges ;  for  the  second,  8  cts.  for  3  :  1  sold  them  at 
the  rate  of  5  oranges  for  14  cts.,  and  gained  52  cts. :  what  num- 
ber did  I  buy  ?  Ans.  240  oranges. 

61.  A  iean  do  a  piece  of  work  in  2^  da.j  B  in  34  da.  :  in  what 
time  can  both  do  it  working  together  ?  Ans.  l|f  da. 


; 


818  RAY'S    PRACTIOAL    ARITHMETIC. 

62.  A  and  B,  at  opposite  points  of  a  field  135  rd.  in  compass, 
start  to  go  round  it  the  same  way,  at  the  same  time :  A  at 
the  rate  of  llrd.  in  2min.,  and  13,  17  rd.  in  3  min.  :  how 
many  rounds  will  each  make,  before  the  one  will  overtake  the 
other?  Ans.  A,  16^  ;  B,  17. 

03.  A  can  mow  a  field  in  3  days,  B  in  5  days,  and  B  and  C 
in  4  days :  in  what  time  can  all  three  together  mow  it  ? 

Ajis.  l|da 

64.  A  and  B  together  can  do  a  piece  of  work  in  15  days;  A 
and  C  in  12  days;  B  and  C  in  10  days :  how  long  will  it  take 
all  together  to  do  it  ?  Ans.  8  da. 

65.  1  mixed  301b.  of  su^ar  at  lOcts.  per  pound;  251b.  at 
12ct8.  per  pound;  41b.  at l5ct8.  per  pound;  and  501b.  at  20 
cts. :  what  is  1  lb.  of  the  mixture  worth  ?         Ans.  1 5^^^^  cts. 

66.  If  10  men  or  18  boys  can  dig  1  acre  in  11  da. ;  find  the 
number  of  boys  whose  assistance  will  enable  5  men  to  dig  6 
acres  in  6  days.  Ans.  189  boys. 

67.  How  many  yd.  of  paper,  24  in.  wide,  will  cover  the  walls 
of  a  room  15  ft.  long,  12  ft.  wide,  10  ft.  high?     Ans.  90  yd. 

68.  A  can  do  a  piece  of  work  in  i  of  a  day ;  B  in  ^  of  a 
day;  and  C  in  a  day  :  in  what  time  can  all  three  do  it  work- 
ing together  ?  Ans.  ^  da. 

69.  If  a  regiment  of  soldiers  be  arranged  in  the  form  of  a 
square,  there  will  be  28  men  on  each  side,  and  50  men  over : 
what  the  whole  number  of  men  ?  Ans.  834  men, 

70.  What  is  the  diagonal  of  a  square,  each  of  the  sides  of 
which  is  40  ft.  ?  Ans.  56.56-Hft 

71.  A  man  after  doing  |  of  a  piece  of  work  in  30  days,  calls 
an  assistant;  both  together  complete  it  in  6  days  :  in  what  time 
could  the  assistant  do  it  alone?  Ans.  21|da. 

72.  A  received  |i  of  an  estate,  and  B  the  remainder ;  A  had 
$7420  more  than  B  :  what  was  the  value  of  the  whole  estate  ? 

An^.  $15900. 

73.  A  room  is  24ft.  long,  18ft.  wide,  and  12ft.  high:  what 
is  the  distance,  measured  diagonally,  from  one  of  the  lower 
corners  to  the  opposite  upper  corner?  Ans.  32.3-|-ft. 

74.  If  1  lb.  of  tea  be  worth  2A  lb.  of  coffee,  and  1  lb.  of  coffee 
be  worth  3^  lb.  of  sugar,  what  will  be  the  value  of  5(>  lb.  of 
tea,  sugar  being  worth  10  cts.  per  lb.  ?  Ans.  $49. 

75.  A,  B,  and  C,  can  together  perform  a  piece  of  work  in  12 
days;  A  alone  can  do  it  in  24  uays ;  B  alone  in  34  davs :  in 
what  time  can  C  do  it  alone  ?  Ans.  81 1  da 


PROMISCUOUS    QUESTIONS.  810 

76.  A,  B,  and  C,  built  a  wall  of  200  feet ;  A  built  as  many 
feet  as  C  and  i  more  ;  B  built  |  as  much  as  A  -.  how  many  feet 
did  each  build  ?  Ans.  A,  80  ft.  ;  B,  60  ft.  ;  C,  60  ft. 

77.  How  many  acres  in  a  square  field,  the  diagonal  being 
42.43  rd.  ?  Ans.  5.62+A. 

78.  A  man  cleared  $19  in  25  days,  by  earning  $1.25  each 
day  he  worked,  and  spending  50  cts.  each  day  he  was  idle  : 
how  many  days  did  he  work?  Ans.  18  da. 

79.  A  can  do  a  job  in  40  da.,  B  in  60  da. ;  after  both  work 
3  da.,  A  leaves  :  when  must  he  return,  that  the  work  may  occu- 
py but  30  da.  ?  Ans.  At  the  end  of  the  13th  da, 

80.  A  gentlemen  left  f^  of  his  estate  to  A:  |  of  the  remain- 
der to  B ;  and  the  rest  to  C;  As  share  was  $1300  more  than 
C's :  what  was  the  share  of  each  ? 

Ans.  A,  $3250;  B,  $3900;  C,  $1950. 

81.  A  cistern  of  500  gal.  has  two  flow  pipes :  one  can  fill 
it  in  3  hours,  and  the  other  in  5  hours  ;  a  third  pipe  being 
added,  the  cistern  was  filled  in  1  hour :  in  what  time  would  the 
third  pipe  fill  it  alone  ?  Ans.  2^  hr. 

82.  I  bought  60  gal.  of  wine  at  $1 .20  per  gal. :  I  sold  20  gaL 
at  $1.50  per  gal.,  and  retained  5  gal.  for  my  own  use  :  at  how 
much  per  gal,  must  1  sell  the  remainder  to  gain  10  %  on  the 
whole  cost?  Ans.  $IA0^ 

83.  I  bought  apples  at  the  rate  of  3  for  4  cts.,  and  sold  them 
at  the  rate  of  2  for  3  cts.  ;  1  cleared  24  cts.  on  the  whole  num- 
ber purchased:  how  many  did  1  buy?         Ans.  144  apples. 

84.  Purchased  a  house  for  $4500 :  it  rented  for  $500  a 
yr. ;  paid  for  insurance  $25;  taxes  1  j-^jj  ^c  ,  and  repairs  $134 : 
what  net  fo  did  it  pay  on  the  investment  ?  Ans.  5|  fc. 

85.  How  many  square  feet  of  flooring  in  a  house  of  4  floors, 
60  ft.  by  30  ft.  within  the  walls,  deducting  from  each  floor  a 
stairway  12  ft.  4  in.  by  8  ft  6  in.  ?  Ans.  6780  sq.  ft.  8^ 

86.  A,  B,  and  C  are  partners:  A  put  in  $700;  B,  $600; 
C,  $400 ;  C's  share  of  the  gain  was  $260  :  what  was  the  whole 
gain?  A71S.  $1105. 

87.  Ten  per  cent  of  120  is  8  less  than  5  ^  of  what  number? 

Ans.  400. 

88.  A  and  B  have  the  same  annual  income  :  A  saves  i  of  his, 
while  B,  who  spends  annually  $350  more  than  A,  at  the  end 
of  four  years,  is  in  debt  $1:^0:  what  is  the  annual  income  of 
each?  A)is>  $960. 


320  RAY'S   PRACTICAL   ARITHMETia 

89.  Two  Globes,  each  5  in.  in  diameter,  and  2  cubes,  each 
5 in.  in  length,  were  melted  into  one  cube:  how  long  was  the 
side  of  this  cube  ?  Ans.  7.24+in. 

90.  Sold  a  cow  for  $25,  losing  16|  ^  ;  bought  another  and 
sold  it  at  a  gain  of  16  ^  ;  1  neither  gained  nor  lost  on  the  two : 
what  the  cost  of  each?  Ans.  Ist,  $30:  2d,  $31.25 

91.  After  a  battle,  in  which  an  army  of  24000  men  were 
engaged,  it  was  found  that  the  number  of  slain  was  ^  of  those 
who  survived ;  the  number  wounded  was  equal  to  ^  of  the 
slain:  find  the  number  wounded.  Ans.  1500. 

92.  I  cut  95  yd  of  cloth  into  3  pieces,  so  that  the  first  was 
three  times  the  second,  and  the  third  one-fourth  the  first:  what 
was  the  length  of  each?       Ans.  1st,  60;  2d,  20;  3d,  15yd. 

93.  A  merchant  purchased  60  yd.  of  cloth,  at  $4  a  yard,  on 
a  credit  of  6  mon. ;  he  sold  it  immediately  for  $250 :  what  did 
he  gain,  money  being  worth  6  ^  ?  Ans.  $16.99-|- 

94.  A  drover  expended  $1500  in  horses,  cows  and  sheep: 
the  horses  cost  $120,  the  cows  $30,  and  the  sheep  ^5  each: 
there  were  six  times  as  many  sheep  as  cows,  and  i  as  many 
horses  as  sheep:  find  the  number  of  each. 

Alls.  10  horses;  5  cows;  30  sheep. 

95.  The  tax  in  a  certain  town  was  1  ct.  6  m.  on  the  dollar,  and 
polls  $1  each ;  A,  who  paid  for  4  polls,  was  charged^328  tax : 
what  the  value  of  his  estate  ?  Ans.  $20250. 

96.  A,  B,  and  C  do  a  piece  of  work  in  a  certain  time :  A  and 
B  together  do  |  of  it ;  B  and  C  together  do  | :  what  part  can  B 
do  alone?  Ans.  |. 

97.  A  and  B  together  received  $1000:  if  B  had  received 
$100  less,  his  share  would  then  have  been  one-half  of  As: 
how  much  did  each  receive  ?         Ans.  A,  $600 ;  B,  $400. 

98.  A  and  B  have  just  $500  :  |  of  A's  money  is  $50  less 
than  I  of  B's  :  what  sum  has  each  ?    Ans.  A,  $200;  B,  j§300. 

99.  If  I  of  the  time  past  noon  is  equal  to  |  of  the  time  to 
midnight,  what  is  the  hour?  Ans.  3  P.  M. 

100.  A  lady  spent  in  one  store  ^  of  all  her  money  and  $1 
more ;  in  another,  ^  of  the  remainder  and  $i  more ;  in  another, 
A  of  the  remainder'and  $1  more;  and  in  another,  ^  of  the  re- 
mainder and  $.\  more;  she  then  had  nothing  left:  what  sum 
had  she  at  first?  Ans.  $20. 


THE   METRICAL    SYSTEM.  821 


APPENDIX. 

THE    METRICAL    SYSTEM 

OF    WEIGHTS    AND   MEASURES. 

333.  The  Metrical  System  is  so  called  because  the 
Meter  is  the  base,  and  the  principal  and  invariable 
unit,  upon  which  the  system  is  founded. 

This  system  was  first  employed  by  the  French ;  and,  hence, 
is  frequently  called  the  French  System  of  Weights  and  Measures. 
Great  confusion  formerly  existed  in  the  weights  and  measures  of 
France.  Each  province  had  its  particular  measures,  which 
caused  great  embarrassment  in  commerce. 

Government  vainly  endeavored  to  establish  a  uniformity,  and 
to  regulate  all  measures  by  those  used  in  Paris. 

In  1790,  the  French  Assembly  proclaimed  the  necessity  of  a 
complete  reform,  and  invited  other  governments  to  join  them  in 
establishing  a  simple  system,  to  be  common  to  all  nations. 

The  cooperation  of  other  nations  could  not  at  the  time  be 
secured,  and  a  commission,  nominated  by  the  Academy  of 
Sciences,  and  composed  of  eminent  scholars,  was  instructed  to 
prepare  a  general  system  of  measures. 

The  new  system  was  adopted,  and  declared  obligatory  after 
Nov.  2,  1801.  But  its  introduction  was  gradual.  It  had  to 
struggle  against  the  local  customs,  and,  for  a  time,  only  in- 
creased the  confusion  by  adding  the  new  measures  to  the  old. 

In  1837,  the  Assembly  enacted  a  law,  rendering  the  exclusive 
use  of  the  new  system  obligatory  after  Jan.  1,  1841;  and  im- 
posed penalties  against  the  further  use  of  the  old  system. 

It  has  since  been  adopted  by  Spain,  Belgium,  and  Portugal,  to 
the  exclusion  of  all  other  weights  and  measures ;  and  is  in  gen- 
eral or  partial  use  in  nearly  all  the  states  of  Europe  and 
America,  and  by  scientific  men  throughout  the  world. 

In  1864,  the  British  Parliament  passed  an  act  allowing  the 
metrical  system  to  bo  used  throughout  the  Empire;  and  in 
3dBk.  21. 


322  ray's  practical  arithmetic. 

1866,  Congress  authorized  its  use  in  the  United  States,  and  pro- 
vided for  its  introduction  into  post-offices  lor  the  weighing  of 
letters  and  papers. 

The  metrical  system,  like  Federal  Money,  is  founded 
on  the  decimal  system  of  notation. 

After  the  principal  unit  of  each  denomination  is 
determined  and  named,  the  names  of  the  higher  de- 
nominations are  formed  by  prefixing  the  Greek  num- 
erals, deka  (10),  hecto  (100),  kilo  (1000),  and  myria 
(10000),  to  the  name  of  the  unit. 

The  names  of  the  lower  denominations  are  formed, 
in  like  manner,  by  prefixing  the  Latin  numerals, 
deci  (.1),  centi  (.01),  and  milli  (.001). 


MEASUEES   OF   LE:N'GTH. 

334.  The  Meter  is  the  unit  for  measure  of  lengths, 
and  is  very  nearly  one  ten -millionth  (tw^'^oooo)  P^^^'^ 
of  the  quadrant  extending  through  Paris  from  the 
equator  to  the  pole. 

Note. — The  meter  is  equal  to  39.37  inches,  nearly,  or  a  little 
less  than  1.1  yards.  It  is  also  nearly  3  feet,  3  inches,  and  3 
eighths  of  an  inch,  which  may  be  remembered  as  the  rule  of  the 
three  threes.     Five  meters  are  about  one  rod. 


TABLE. 

10  Meters,  marked  ]M 

—  1  Dekameter, 

marked  D.  M. 

1 0  Dckametors    .     . 

=  1  Hectometer, 

"      n.  M. 

10  Hectometers  .     . 

—  1  Kilometer, 

"       K.  M. 

10  Kilometers     .     . 

—  1  Myriameter, 

"       M.  M. 

Also, 

-jJg  of  a  Meter      .     . 

—  1  Decimeter, 

marked   d.  m. 

Jg  of  a  Decimeter  . 

—  1  Centimeter, 

c.  m. 

•Ai  of  a  Centimotcr  . 

— 1  Millimeter, 

"       m.  m. 

THE    METRICAL    SYSTEM. 


823 


o 


to- 


co- 


tfk. 


00- 


z 

en 


The  figure  in  the  margin  rep- 
resents the  exact  length  of  a 
decimeter,  divided  into  10  equal 
parts  or  centimeters,  and  these 
subdivided  into  10  equal  parts, 
or  millimeters. 

The  adjoining  figure  repre- 
sents a  scale  of  four  inches. 

This  illustration  will  aid  the 
pupil  in  gaining  a  clearer  idea 
of  the  relative  length  of  the 
different  units  of  measure  of  the 
two  systems. 

Meters  are  usually  separated 
from  the  lower  denominations 
by  the  decimal  point.  Thus,  7 
meters,  3  decimeters,  and  5  cen- 
timeters, are  written  7.35  ]\I.; 
read,  7  meters  and  85  centi- 
meters. 

G-reat  distances  are  reckoned 
by  kilometers ;  thus,  the  length 
of  a  canal  may  be  given  as  48 
kilometers,  7  hectometers ;  writ- 
ten, 48.7  K.  M.  In  like  man- 
ner, if  the  distance  between  two 
cities  is  32  myriameters,  5  kilometers,  it  is  written, 
325.  K.  M. 

Very  small  distances  are  reckoned  by  millimeters ; 
for  examj^le,  if  the  thickness  of  a  piece  of  glass  is  8 
millimeters  and  5  tenths,  it  is  written,  8.5  m.  m. 

Millimeters,  centimeters,  meters,  and  kilometers, 
arc  the  denominations  most  used ;  decimeters,  deka- 
meters,   hectometers,  etc.,  are   not   used   in   ordinary 


824  ray's  practical  arithmetic. 

transactions ;  as  in  Federal  money,  the  denominations 
of  eagles,  dimes,  and  mills  are  unused. 

Example. — How  many  millimeters  in  4.27  M.? 

Solution.— Since  1  M.  is  equal  to  1000  operation 

m.  m.,  4.27  M.  will  be  equal  to  4.27X1000        497' 
m.  m.,  which  is  4270  m.  m.  "^1  000 

Conclusion. — Therefore,  4.27  M.  equals        4270.00  m  m 
4270  millimeters.     Hence,  the 

Rule  for  Reduction  Descending. — Multiply  the  given 
quantity  by  that  number  of  the  lower  denomination  which 
makes  a  unit  of  the  required  denomination. 

XoTE. — Since  the  multiplier  is  always  10,  100,  1000,  etc.,  the 
operation  may  be  performed  by  removing  the  point  as  many 
places  to  the  right  as  there  arc  ciphers  in  the  multiplier. 

Example. — How  many  kilometers  in  36429  M.? 

Solution. — Since  1000  meters  = 
1   kilometer,  36429  M.  ^  as  many  operation. 

kilometers,    as    1000    is    contained         1000)36429 
times   in   36429;    36429-^1000=  "36~429K  M 

36.429  K.  M. 

Conclusion.— Therefore,  36429  M.  equal  36.429  K.  M. 
Hence,  the 

Rule  for  Reduction  Ascending.— Di'yz^e  the  given 
quantity  by  that  number  of  its  own  denomination  which 
makes  a  unit  of  the  required  denomination. 

Note. — Since  the  divisor  is  always  10,  100,  1000,  etc.,  the 
operation  may  bo  performed  by  removing  the  decimal  point  as 
many  places  to  tlio  left  us  there  are  ciphers  in  the  divisor. 


THE    METRICAL    SYSTEM.  325 

EXAMPLES   FOR   PEACTICE. 

1.  Reduce  30.75  M.  to  centimeters.  Ans.  3075. 

2.  Reduce  4.5  K.  M.  to  meters.  Ans.  4500. 

3.  Reduce  75  m.  m.  to  meters.  Ans.  .075. 

4.  Reduce  25  D.  M.  to  decimeters.  Ans.  2500. 

5.  Reduce  35  M.  M.  to  hectometers.  Ans.  3500. 

6.  Reduce  36.5  M.  to  dekameters.  Ans.  3.65. 

7.  Reduce  36.5  M.  to  decimeters.  Ans.  365. 

MEASURES   OF   SURFACE. 

335.  The  Square  Meter  is  the  unit  of  Square 
Measure.  It  is  a  square  of  which  each  side  is  a 
meter. 

The  square  meter  is  equal  to  100  square  decimeters; 
for,  if  a  square  is  constructed,  each  side  of  which  is 
1  meter,  or  10  decimeters  long,  the  area  will  be  10  X 
10=100  square  decimeters;  hence,  one  square  deci- 
meter equals  jjg  of  a  square  meter. 


TABLE. 

100  Sq. 

Meters,  (M.2) 

—  1   Sq.  Dekameter,  marked  D.  M.2 

100   " 

Dekameters, 

=  1    "    Hectometer,       "       H.M.2 

100   " 

Hectometers, 

=:1    "    Kilometer,         "       K.  M.2 

100   " 

Kilometers, 

=  1    "    Myriameters,     "       M.  M.2 

1^15  Sq. 

Meter  .     .     . 

—  1   Sq.  Decimeter,  marked   d.  m,^ 

1        u 
TUB 

Decimeter 

—  1    "    Centimeter,       "         c.  m.2 

j^(j    "    Centimeter    .     =1    "    Millimeter         "       m.  m.^ 

Note. — l^ince  100  units  of  each  order  make  one  of  the  next 
higher,  each  order  should  occupy  hvo  places.  Thus,  534.260780 
M.^  signifies  5  square  decimeters,  34  square  meters,  26  square 
decimeters,  7  square  centimeters,  and  80  square  millimeters; 


326  ray's  practical  arithmetic. 

and  may  be  read,  five  hundred  and  thirty-four  square  meters, 
two  hundred  and  sixty  thousand,  seven  hundred  and  eighty 
square  millimeters. 


EXAMPLES  FOE  PKACTICE. 

1.  Eeduce  26304  d.  m.2  to  M2.  Ans.  263.04. 

2.  Eeduce  1460  m.  m.2  to  M^.  Ans.  .00146. 

3.  Eeduce  14  D.  M.2  to  d.  m2.  Ans.  140000. 

4.  Eeduce  20  M.  M.2  to  D.  M^.  Ans.  20000000. 

5.  Eeduce  3294  M.-  to  H.  M^.  Ans.  .3294. 

6.  Eeduce  14.1  H.  M.2  to  M2.  Ans.  141000. 

336.  In  measuring  land,  only  three  of  the  denom- 
inations of  square  measure  are  used,  and  these  receive 
shorter  names. 

The  Are,  or  square  dckameter,  is  the  unit  of  land 
measure. 

Note. — The  Are  is  equal  to  119.6  square  yards.  The  Are 
equals  100  square  meters;  the  Centare,  1  square  meter;  and 
the  Hectare,  10000. 

TABLE. 

100  Ares,  (A.)  .     .     =1  Hectare,  marked  II.  A. 
Also, 

j^Q  Are,   ....     =1  Centare,  marked  c.  a. 

EXAMPLES    FOE   PEACTICE. 

1.  Eeduce  26.25  A.  to  centarcs.  Ans.  2625. 

2.  Eeduce  21  II.  A.  to  ares.  Ans.  2100. 

3.  Eeduce  3.5  H.  A.  to  M2.  Ans.  35000. 

4.  Eeduce  14209  M.2  to  ares.  Ans.  142.09. 

5.  Eeduce  3.8  A.  to  square  meters.  Ans.  380. 
G.  Eeduce  27  d.  ni.2  to  contarus.  Ans.  .27. 


THE    METRICAL    SYSTEM.  327 

CUBIC   MEASURE. 

337.  The  Stere,  or  cubic  meter,  is  the  unit  of 
cubic  or  solid  measure. 

The  cubic  meter  equals  1000  cubic  decimeters;  for, 
if  a  cube  be  taken,  each  side  of  which  is  one  meter 
or  ten  decimeters,  the  cubic  contents  Avill  equal  10  X 
10x10=1000  cubic  decimeters;  therefore,  one  cubic 
decimeter  equals  y^jVir  ^^  ^  cubic  meter.  , 

Note. — A  cubic  meter  equals  1.308  cubic  yards. 

TABLE. 

j^L_  Cubic  M.,  (M.3)  =1   Cubic   d.  m  ,     marked    d.  m.^ 
_^i__      "      d.  m..     .     =1        "        cm.,  "  cm.  3 

yJjj^      "      cm.     .     =1       "       m.  m.,  "        m.  m.^ 

All  solid  bodies,  except  wood,  are  measured  by  cubic 
meters,  and  the  lower  denominations. 

For  the  measurement  of  fire-wood  and  building 
timber,  use  the  following 

TABLE. 

10  Steres,  (S.)  .     .     =1   Dekastere,  marked  D.  S. 
Also, 

y'q  Stere    ....     =1   Decistere,     marked    d.  s, 

EXAMPLES   FOR   PRACTICE. 

1.  Reduce  9  S.  to  decisteres.  Ans.  90. 

2.  Reduce  4.19  d.  s.  to  dekasteres.         Ans.  .0419. 
8.  Reduce  29  M.3  to  c.  m.3  Ans.  29000000. 

4.  Reduce  25  D.  S.  to  decisteres.  Ans.  2500. 

5.  Reduce  16.32  M.^  to  dekasteres.       Ans.  1.632. 

6.  Reduce  14000  c.  m.^  tv^  cubic  meters.    Ans.  .014. 


328 


RAY  S    PRACTICAL    ARITHMETIC. 


WEIGHTS. 

838.  The  Gram  is  the  unit  of  weight, 
in  weighing  all  substances. 


It  is  used 


Note. — The  gram  is  equal  to  15.432  Troy  grains,  which  is  the 
weight,  in  vacuum,  of  a  cubic  centimeter  of  water,  at  39.2° 
Fahrenheit  At  this  temperature  water  attains  its  greatest 
density.  / 


Also, 


10  Grams,  (G.) 
10  Dekagrams 
10  Hectograms 


■J^  Gram 
-j-'g  Decigram 
-Jg  Centigram 


TABLE.  • 

:1   Dekagram,  marked  D.  G. 

:  1  Hectogram,  "  H.  G. 

:1   Kilogram,  "  K.  G. 

:1   Decigram,  marked  d.  g, 

■  \   Centigram,  "  c.  g. 

\  Milligram,  "  m.  g. 


The  gram  and  its  subdivisions  are  used  in  mixing 
and  compounding  medicines,  and  in  all  cases  where 
great  exactness  is  required. 

The  new  five-cent  coin,  made  of  nickel  and  copper, 
and  dated  1866,  is  20  millimeters  in  diameter,  and 
weighs  5  grams. 

The  kilogram  is  the  principal  and  ordinary  unit  of 
weight,  and  is  a  little  more  than  2-^  pounds  avoirdu- 
pois. 

There  arc  two  other  denominations  employed  in 
weighing  heavy  articles,  the  quintal,  which  equals  100 
kilograms;  and  the  tonneau,  which  equals  1000  kilo- 
grams. 

The  tonneau  is  greater  than  the  short  ton  of  2000 
pounds,  and  less  than  the  long  ton  of  2240  pounds. 
Its  exact  weight  is  2204.6  pounds  Avoirdupois. 


THE    METRICAL    SYSTEM. 


329 


EXAMPLES   FOE  PEACTICE 


1.  Eeduce 

2.  Eeduce 

3.  Eeduce 

4.  Eeduce 

5.  Eeduce 

6.  Eeduce 

7.  Eeduce 

8.  Eeduce 

9.  Eeduce 


1428.06  G.  to  kilograms. 
2.8  K.  G.  to  grams. 
119  H.  G.  to  decigrams. 
171300  G.  to  quintals. 
3600  G.  to  kilograms. 
19200  m.  g.  to  grams. 
4  quintals  to  grams. 
29  G.  to  centigrams. 
492  D.  G.  to  quintals. 


^7is.  1.42086. 

Ajis.  2800. 

Alls.  119000. 

Ans.  1.713. 

A71S.  3.6. 

Ans.  19.2. 

Ans.  400000. 

Ans.  2900. 

Ans.  .0492. 


MEASUEES   OF   CAPACITY. 

S30.  The  Liter  is  the  unit  of  capacity.  It  is  equal 
to  a  cubic  decimeter. 

Note. — The  liter  equals  1.0567  quarts,  U.  S.  Wine  measure. 
The  denominations  most  used  in  this  table  are  liter  and  hecto- 
liter. 


AlsOj 


10  Liters,  (L) 
10  Dekaliters 
10  Hectoliters 

J^  Liter  .  . 
J|)  Deciliter  . 
Jq  Centiliter    . 


TABLE. 

:1  Dekaliter, 
l   Hectoliter, 
:i   Kiloliter, 

=1  Deciliter, 
=1  Centiliter, 
-A  Milliliter, 


marked    D.  L. 
"  H.  L 

"  K.  L. 


marked     d.  L 

c.  L 

"         m.  1. 


The  liter  is  equal  to  the  cubic  decimeter ;  the  milliliter 
being  the  one-thousandth  part  of  a  liter,  equals  a  cubic  centi- 
meter, and  the  kiloliter  equals  a  cubic  meter.  The  kiloliter  or 
cubic  meter  of  water  weighs  one  tonneau;  the  liter  of  water 
weighs  a  kilogram;  and  the  milliliter  of  water  weighs  a 
gram. 


330 


KAY  S    PRACTICAL    AllITUMETIC. 


EXAMPLES    FOE  PEACTICE. 


1.  Beduce  24  L.  to  centiliters. 

2.  Reduce  183  ILL.  to  deciliters. 

3.  Eeduce  .02345  K.  L.  to  liters. 

4.  Eeduce  1.4  L.  to  milliliters. 

5.  Reduce  3  'M^  to  liters. 

6.  Reduce  40360  L,  to  hectoliters. 

7.  Reduce  1400  L.  to  cubic  meters 


Ans.  2400. 

Ans.  183000. 

.4ns.  23.45. 

.4ns.  1400. 

Ans.  3000. 

^ns.  403.6. 

Ans.  1.4. 


340.  In  order  that  the  pupil  may  become  thor- 
oughly acquainted  with  the  value  of  the  different 
denominations  of  the  Metrical  System,  those  in  ordi- 
nary use  are  presented  in  the  following  Tables  : 

MEASURES  OF  LENGTH. 


NAMES. 

Value  in  Meters. 

Value  in  ordinary  denominations. 

Mvriameter. 

10000. 

6.2137  miles. 

Kilometer. 

1000. 

.62137  mile. 

Hectometer. 

100. 

328-J2  feet. 

Dckameter. 

10. 

393.7  inches. 

Meter. 

1. 

39.37  inches. 

Decimeter. 

.1 

3.937  inches. 

Centimeter 

.01 

.3937  inch. 

Millimeter 

.001 

.0394  ineli. 

MEASURES  OF  SURFACE. 


NAMES. 

Value  in  M^ 

Value  in  ordinary  denominations. 

Hectare. 

Are. 

Centanv 

10000. 
100. 

1. 

2.471  acres. 

119.6  sq.  yds. 

15r)0  sq.  in 

TUE    METRICAL    SYSTEM. 


331 


MEASURES    OF   CAPACITY. 


NAMES. 

Value  in  Liters. 

in  Dry  Mea're. 

in  Wine  Mea're. 

Kiloliter,  or  Stere. 

1000. 

1.308  cu.  yds. 

264.17  gal. 

Hectoliter. 

100. 

2.8375  bush. 

26.417  gal. 

Dekaliter. 

10. 

9.08  qt. 

2.6417  gal. 

Liter. 

1. 

.908  qt. 

1.0567  qt. 

Deciliter. 

.1 

6.1022  cu.  in 

.845  gill. 

Centiliter. 

.01 

.6102  cu.  in. 

.338  fluid  oz. 

Milliliter. 

.001 

.061  cu.  in. 

.27  fluid  dra'm 

WEIGHTS. 


NAMES. 

Value  in 
Grams. 

Quan'y  of  Water. 

Ordinary 

weights. 

Tonneau. 

1000000. 

lM3,orl  K.L. 

2204.6  lb. 

Av'pois. 

Quintal 

100000. 

1  Hectoliter. 

220.46  lb. 

u 

Mvriagram. 

10000. 

10  Liters. 

22.046  lb. 

u 

Kilogram. 

1000. 

1  Liter. 

2.2046  lb. 

u 

Hectogram. 

100. 

1  Deciliter. 

3.5274  oz. 

l( 

Dekagram. 

10. 

10  cm. 3 

.3527  oz. 

u 

Gram. 

1. 

1  c.  m.  3 

15.432  gr. 

Troy. 

Decigram. 

.1 

y'o  cm. 3 

1.5432  gr. 

u 

Centigram. 

.01 

10  m.  m.3 

.1543  gr. 

u 

Milligram. 

.001 

1  m.  m.  3 

.0154  srr. 

1( 


341.    MISCELLANEOUS    EXAMPLES. 

1.  The  new  5-cent  piece  weigh.s  5  grams :  how  much 
will  100  such  pieces  weigh?  Ans.  500  Gr. 

2.  Ten  liters  of  a  certain  liquid  weigh   92  K.  G. : 
what  is  the  weight  of  a  deciliter?  Ans.  920  G. 

3.  One  hectogram  of  goods  costs  $5.35  :  what  costs 
one  kilogram?  Ans.  $53.50. 


332  bay's  practical  arithmetic. 

4.  One  decistere  of  wood  is  worth  28  cents:  what 
is  a  stere  worth?  Ajis.  §2.80. 

5.  A  hectoliter  of  wheat  costs  §6.25:  what  is  the 
price  of  a  dekaliter?  Ans.  $.625. 

6.  A  hectoliter  of  wine  costs  §2-5.10  :  what  is  the 
price  of  a  liter?  Ans.  §.251. 

7.  A  kilogram  of  wool  costs  §1.875:  what  will 
be  the  cost  of  100  kilograms,  or  1  quintal  ? 

Ans.  §187.50. 

8.  A  liter  of  wine  weighs  880  G. :  what  is  the 
weight  of  a  hectoliter?  Ans.  88  K.  G-. 

9.  A  wine  merchant  sold  1270  liters,  487  liters, 
1563  liters,  and  2345  liters:  how  many  hectoliters 
did  he  sell?  Ans.  56.65  ILL. 

10.  A  vase  weighing  24.67  K.  G.,  contains  18.79 
K.  G.  of  liquid :  what  is  the  weight  of  the  empty 
vase?  Ans.  5.88  K.  G. 

11.  A  courier  traveled  135  K.  M.  in  one  day:  how 
far  can  he  travel  in  45  days  ?  Ans.  6075  X.  M. 

12.  How  much  will  135.60  M.  of  cloth  cost  at 
$1.16  a  meter?  Ajis.  §157.296. 

13.  Bought  25  hogsheads  of  wine,  of  225  liters 
each,  at  the  rate  of  §.156  a  liter:  how  much  did  it 
cost?  Ans.  §877.50. 

14.  AVhat  is  the  cost  of  21  pieces  of  cloth  of  42  M. 
each,  at  §5.69  a  meter?  Ans.  §5018.58. 

15.  A  man  bought  3.80  M.  of  velvet  at  §2.16  a 
meter:  how  much  did  it  cost  him?  Ans.  §8.208. 

16.  What  will  be  the  price  of  5  deciliters  of  wine 
at  28  cents  a  liter?  Ans.  §.14. 

17.  What  Avill  be  the  cost  of  45  IL  A.  of  land  at 
§3.32  an  are?  A?is.  $14940. 

18.  What  is  the  price  of  15  L.  of  wine,  if  it  costs 
$14.06  a  hectoliter?  Ans.  §2.109. 


TUB    METRICAL    SYSTEM.  333 

19.  A  merchant  paid  $457.92  for  cloth  at  $3  a 
meter :  how  many  meters  did  he  buy?   Ans.  152 .  64  M. 

20.  A  man  traveled  665  K.  M.  in  7  days:  how  far 
did  he  go  each  day?  Ans.  95  K.  M. 

21.  A  manufacturer  bought  380  steres  of  wood  for 
$454.10:  how  much  was  that  a  stere?     Ans.  $1,195. 

22.  If  235  L.  of  wine  cost  $34,545,  what  is  the 
cost  of  one  liter?  Ans.  $.147. 

23.  A  tower  is  142.695  M.  high  ;  it  has  453  steps: 
what  the  height  of  each?  Ans.  .315  M. 

24.  What  is  the  cost  of  248.35  31.  of  gold  thread, 
at  the  rate  of  $1.58  a  meter?  Ans.  $392,393. 

25.  If  1  L.  of  a  liquid  weighs  1.25  K.  G.,  how 
much  will  25.8  L.  weigh?  .4ns.  32.25  K.  G. 

26.  I  bought  846.75  K.G.  of  coffee  at  56  cents  a 
kilogram:  what  did  it  cost  me?  Ans.  $194.18. 

27.  A  traveler  has  to  go  548  K.  M.  in  14  days: 
how  far  ought  he  to  go  in  one  day  ?    Ans.  39.14-f-K.M. 

28.  What  must  be  paid  for  367  S.  of  wood,  at 
$3,175  a  stere?  Jns.  $1165.225. 

29.  How  many  liters  in  9684  flasks,  each  contain- 
ing .25  L.?  Ans.  2421  L. 

30.  How  many  hectoliters  of  oats  in  4685  sacks, 
if  each  contains  1  H.  L.  6  D.  L.  ?         Ans.  7496  H.  L. 

31.  If  4957  loaves  of  sugar  weigh  15684.08  K.  G., 
what  is  the  weight  of  one  loaf?     Ans.  3.164  -f  K.G. 

32.  What  will  be  the  price  of  a  hectogram  of 
sugar,  if  a  loaf  weighing  8  K.  G.,  6  D.  G.,  and  6  G., 
costs  $32.95  ?  Ans.  $.408  +. 

33.  How  much  cloth  1.85  M.  wide  will  it  take  to 
line  a  cloak,  made  of  6.5  M.  of  cloth  1.25  M.  wide? 

Am.  4.39 +  M. 

34.  A  silver  half-dollar  weighs  12.441  G. :  what  is 
the  weightof$11.50  in  half  dollars?   Ans.  286.143G. 


334  ray's  practical  arithmetic. 

35.  If  a  meter  of  cloth  costs  $2.50,  how  much  can 
be  purchased  for  $72.10  ?  Aris.  28.84  M. 

36.  "What  is  the  cost  of  a  field  of  12  H.  A.,  if  an. 
are  costs  §6.35?  Ans.  §7620. 

37.  What  is  the  cost  of  a  field  containing  3  A.,  at 
the  rate  of  $.625  a  square  meter?         Ajis.  $187.50. 

38.  If  3  H.  L.  of  wine  cost  $11  J,  what  is  the  price 
of  1  liter?  A?is.  $.137  -f. 

39.  "VYhat  quantity  of  wheat  is  contained  in  792 
sacks,  each  containing  1  H.  L.  and  2  D.  L.? 

.4ns.  950.411.  L. 

40.  A  has  a  garden  of  3.6  II.  A. ;  he  sold  30  31^ 
to  a  neighbor:  how  much  had  he  left?    Ans.  3.597  H.A. 

41.  A  block  of  marble  .72  M.  long,  .48  M.  wide, 
and  .5  M.  thick,  costs  $.864:  what  is  the  cost  of  one 
cubic  meter?  Ayis.  $5. 

342.    REDUCTION. 
Example. — Reduce  1  mile  to  meters. 

OPERATION. 

Solution.— First,  reduce      39.37)63360(1609.347+ H. 

the  1  mile  to  inches.     It  3937 

equals  633G0  inches;  and  qqgqq 

since  39.37   inches  make  23622 
one  meter,  63360   inches 
will  make  as  many  meters 

as  39.37  is  contained  times  

in   63360  =  1609.347  +  13670 

meters.  11811 


36800 
35433 


rr.,  r  18590 

Conclusion. — Thororore,  15748 

1  mile  equals  1609.347  +  

meters.     Hence,  the  *-o4-.U 

27559 


THE    METRICAL    SYSTEM.  335 

Rule  for  reducing  the  ordinary  denominations  to 
the  metrical  system, — Divide  by  the  number  of  the  given 
denomination  making  one  of  the  required  metrical  denom- 
ination. 

Example. — Eeducc  2000  M.  to  yards. 

Solution. — Since  I  meter  equals  39.  operation, 

37  inches,  2000  meters  equal    2000X  39.37 

39.37  inches  =  78740   inches      which,  2000 

reduced  to  yards,  equal  2187  yd.  8  in,  12')78740  00  in 

Conclusion. —  Therefore,     2000    M.  3)6561  ft.  8  in. 

equal  2187  yd.  8  in.     Hence,  the  ^^  ^^ 

^  ^  2187  yd. 

Rule  for  reducing'  metrical  denominations  to  the 
ordinary  system. — Multiply  by  the  number  expressing 
the  equivalent  of  the  unit  of  the  given  metrical  denom- 
ination. 

EXAMPLES   FOR  PRACTICE. 

1,  Eeduce  8  inches  to  centimeters, 

Ans.  20.32 -fern. 

2,  Reduce  25  feet  to  meters.  Ans.  7.62  -f  M. 

3,  How  many  K.  M.  from  Cincinnati  to  Dayton, 
the  distance  being  60  miles?  Ans.  96.56  K,  M. 

4,  Mt.  Everest,  the  highest  mountain  in  the  world, 
is  29100  ft.  high;  what  is  its  height  in  meters? 

Ans.  8869.69.  M. 
6.  Reduce  20  K.  M.  to  miles.      Ans.  12.427  +  mi. 

6.  Reduce  49  M.  to  ordinary  denominations. 

Ans.  9rd.  4  yd,  3.13  in. 

7.  A    merchant    in    Holland    bought    360  M.    of 
linen:  how  many  yards  did  he  buy?     Ans.  393.7yd. 

8.  How  many  hectares  in  a  quarter  section  (160 
acres)  of  land?  Ans.  64.74  +  H.  A. 


336  ray's  practical  arithmetic. 

9.  How  many  square  yards  in  a  roll  of  paper,  9 
M.  long  and  .5  M.  wide?  Ans.  5.38  -f  sq.  yd. 

10.  Eeduce  2^  bushels  to  liters.         Ans.  88.1  +  L. 

11.  Reduce  32  L.  to  gallons.  Ans.  8.45  -|-  gal. 

12.  How  many  hectoliters  in  a  barrel  containing  42 
gallons  of  vinegar?  Ans.  1.5899  -j-  H.L. 

13.  A  grocer  sold  4  lb.  of  tea:  how  much  would  it 
have  weighed  in  metrical  denominations? 

Ans.  1.814 +  K.  G. 

14.  A  load  of  hay  weighing  .92  tonneau,  was  sold 
at  1  cent  a  pound,  Avoirdupois :  for  how  much  did  it 
sell?  Ans.  $20.28-}- 

15.  A  has  Manila  cordage  which  he  sells  at  26  ct. 
a  pound:  how  shall  he  mark  it  to  sell  by  the  kilo- 
gram? Ans.  57  cts. 

16.  B  has  sugars  marked  at  14,  16,  and  20  ct.  a 
pound :  what  is  the  price  of  each  by  the  kilogram  ? 

Ans.  31,  35,  and  44  cts. 

17.  Bought  36  lb.  of  tobacco  for  §28.80:  at  how 
much  per  K.  G.  shall  I  sell  it  to  gain  $1.44? 

Ans.  $1.85  -f- 

18.  A  single  letter  is  allowed  by  law,  to  weigh  15 
grams:  how  many  such  letters  in  a  mail-bag,  weigh- 
ing 41  lb.  5  oz.,  if  the  bag  alone  weighs  8  lb.  3  oz. 
15.2  dr.?  ^       Ans.  1000. 

19.  A  man  bought  25  lb.  of  tea  at  $1.80  a  pound; 
he  exchanged  it  for  five  times  its  weight  in  coffee, 
which  he  sold  at  $.86  a  kilogram:  did  he  gain  or 
lose  by  the  bargain,  and  how  much? 

Ans.  Gained,  $3.76  -\-, 


THE     END. 


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POLITICAL  ECONOMY. 


ANDREWS  S  MANUAL  OF  THt  CONSTITUTION. 

Manual  of  the  Constitut'on  of  the  United  States.  De- 
signed for  the  Instruction  of  American  Youth  in  the  Du- 
ties, Obligations  and  Rights  of  Citizenship.  By  Israel 
Ward  Andrews,  D.  D.  ,  President  of  Marietta  Ccllr^f. 
i2mo.  cloth,  408  pp. 

While  the  primary  object  has  been  to  provide  a  suitable  text-book. 
a  conviction  that  a  knowledge  of  our  government  can  not  be  too  widely 
diffused,  and  that  large  numbers  would  welcome  a  good  book  on  this 
subject,  has  led  to  the  attempt  to  make  this  volume  a  manual  adapted 
for  consultation  and  reference,  as  well  for  citizens  at  large  as  for  stu- 
dents. With  this  end  in  view  the  work  embodies  that  kind  of  informa- 
tion on  the  various  topics  which  an  intelligent  citizen  would  desire  to 
possess. 
GREGORYS  POLITICAL  ECONOMY. 

A  New  Political  Economy.  By  John  M.  Gregory, 
LL.D.,   late  President ,  III   Jndustrial  University.      i2mo, 

393  PP- 

An  essentially  new  stiitemcnt  of  the  facts  and  pnncipiei^  oi  ioiitical 
Fconomy,  in  the  following  particulars: 

I  Tnr-  clear  recognition  o  the  three  freat  economic  facts  of  Wants, 
Work  and  Wealth  as  the  principal  and  constant  factors  of  the  indus- 
tries, and  as  constituting,  ihc-ri-fore,  the  field  of  Economic  Science. 

-  The  ,':cognition  of  man  and  of  the  two«ffreat  crystallizations  cf 
man  into  soci^tj;  anil  i.-.'<  states,  as  presenting  three  distinct  fields  of 
Economic  Scie!;-e,  each  having  its  own  set  of  problems,  and  each  iis 
own  species  of  c.nntilies  or  factors,  to  be  taken  into  account  in  the 
solution  of  problems. 

3.  A  new  definitit^n  ;i;»d  description  cf  Value  as  made  up  of  its  three 
essential  and  c^er-prebv.nt  factoi--  forming  the  triangle  of  Value,  and 
evidenced  by  iSe  (-ie-^r  explan>.  ion  thry  afford  of  the  various  fluctua- 
tions of  prices. 

4.  The  new  divlsion'and  distribution   of  th( 
of  these  nev    i-mdament  il  futr  and- definitions. 

5.  The  aid  tendered  to  th<    nr.dcr  rm  "'     t..ri,  ,.■   \s    x^h,- (Vr.nyT.m^  .\t\A 

synoptical  vicvs. 

Van  ANT^^'pr,  pR4nn  ,%  He  PiiMis 

CIN    ",^NA.TI  fn.i  NEW 


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